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Mathematics-II (MATH F112)Linear Algebra
Jitender Kumar
Department of MathematicsBirla Institute of Technology and Science Pilani
Pilani-333031
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 1 / 122
Chapter: 5 (Linear Transformations)
1 Introduction to Linear Transformations2 The Dimension Theorem3 One-to-One and Onto Linear Transformations4 Isomorphism5 Coordinatization (4.7)6 The Matrix of a Linear Transformation
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Section 5.1: Linear Transformations
Let V and W be real vector spaces.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 3 / 122
Section 5.1: Linear Transformations
Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 3 / 122
Section 5.1: Linear Transformations
Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT) if and only if both of thefollowing are true:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 3 / 122
Section 5.1: Linear Transformations
Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT) if and only if both of thefollowing are true:
L(u + v) = L(u) + L(v) for all u, v ∈ V
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 3 / 122
Section 5.1: Linear Transformations
Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT) if and only if both of thefollowing are true:
L(u + v) = L(u) + L(v) for all u, v ∈ V
L(cu) = cL(u) for all c ∈ R and all u ∈ V
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 3 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Check whether L is a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Check whether L is a LT.
Solution: Let A, B ∈ Mmn and c ∈ R. Note that
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Check whether L is a LT.
Solution: Let A, B ∈ Mmn and c ∈ R. Note that
L(A + B) = (A + B)T = AT + BT = L(A) + L(B)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Check whether L is a LT.
Solution: Let A, B ∈ Mmn and c ∈ R. Note that
L(A + B) = (A + B)T = AT + BT = L(A) + L(B)
L(cA) = (cA)T = cAT = cL(A).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 1: For A ∈ Mmn, consider the mapping
L : Mmn → Mnm given by
L(A) = AT .
Check whether L is a LT.
Solution: Let A, B ∈ Mmn and c ∈ R. Note that
L(A + B) = (A + B)T = AT + BT = L(A) + L(B)
L(cA) = (cA)T = cAT = cL(A).
Hence, L is a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 4 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider
L(2([1, 2])) = L([2, 4])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider
L(2([1, 2])) = L([2, 4])
= [2, 4, 8] 6= 2L([1, 2])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider
L(2([1, 2])) = L([2, 4])
= [2, 4, 8] 6= 2L([1, 2])
Thus, L(c([x, y])) 6= cL([x, y]) ∀c ∈ R and [x, y] ∈ R2
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Example 2: For each [x, y] ∈ R2, consider a map
L : R2 → R
3 given by
L([x, y]) = [x, y, xy].
Check whether L is a LT.
Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider
L(2([1, 2])) = L([2, 4])
= [2, 4, 8] 6= 2L([1, 2])
Thus, L(c([x, y])) 6= cL([x, y]) ∀c ∈ R and [x, y] ∈ R2
Hence, L is not a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 5 / 122
Exercise: Check which of the following maps are LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Exercise: Check which of the following maps are LT.1 L : P2 → R
3 given by L(a + bx + cx2) = [a, b, c].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Exercise: Check which of the following maps are LT.1 L : P2 → R
3 given by L(a + bx + cx2) = [a, b, c].2 L : R
3 → R2 given by L([x, y, z]) = [x − y, y + z].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Exercise: Check which of the following maps are LT.1 L : P2 → R
3 given by L(a + bx + cx2) = [a, b, c].2 L : R
3 → R2 given by L([x, y, z]) = [x − y, y + z].
3 L : R2 → R
2 given by L([a, b]) = [a,−b].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Exercise: Check which of the following maps are LT.1 L : P2 → R
3 given by L(a + bx + cx2) = [a, b, c].2 L : R
3 → R2 given by L([x, y, z]) = [x − y, y + z].
3 L : R2 → R
2 given by L([a, b]) = [a,−b].4 L : R → Φ given by L(x) = sinx.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Exercise: Check which of the following maps are LT.1 L : P2 → R
3 given by L(a + bx + cx2) = [a, b, c].2 L : R
3 → R2 given by L([x, y, z]) = [x − y, y + z].
3 L : R2 → R
2 given by L([a, b]) = [a,−b].4 L : R → Φ given by L(x) = sinx.5 L : R → R given by L(x) = x2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 6 / 122
Linear Operator: Let V be a vector space.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 7 / 122
Linear Operator: Let V be a vector space. A linearoperator on V is a LT whose domain and codomainare both V.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 7 / 122
Linear Operator: Let V be a vector space. A linearoperator on V is a LT whose domain and codomainare both V.
Example 3: The mapping L : R3 → R
3 given byL([x, y, z]) = [x, y,−z] is a linear operator.
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Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT.
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Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 8 / 122
Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
1 L(0V) = 0W
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Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
1 L(0V) = 0W
2 L(−v) = −L(v), for all v ∈ V
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 8 / 122
Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
1 L(0V) = 0W
2 L(−v) = −L(v), for all v ∈ V3 For n ≥ 2 and a1, a2, . . . , an ∈ R,
If v = a1v1 + a2v2 + · · · + anvn, then
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 8 / 122
Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
1 L(0V) = 0W
2 L(−v) = −L(v), for all v ∈ V3 For n ≥ 2 and a1, a2, . . . , an ∈ R,
If v = a1v1 + a2v2 + · · · + anvn, thenL(v) = L(a1v1 + a2v2 + · · · + anvn)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 8 / 122
Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then
1 L(0V) = 0W
2 L(−v) = −L(v), for all v ∈ V3 For n ≥ 2 and a1, a2, . . . , an ∈ R,
If v = a1v1 + a2v2 + · · · + anvn, thenL(v) = L(a1v1 + a2v2 + · · · + anvn)
= a1L(v1) + a2L(v2) + · · · + anL(vn).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 8 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]
L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]
L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])
= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]
L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])
= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])
= −3[−2, 1, 0] + 2[3,−2, 1] + 4[0,−1, 3]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Example 4: Let L : R3 → R
3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].
Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R
3.
Solution:
[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]
L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])
= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])
= −3[−2, 1, 0] + 2[3,−2, 1] + 4[0,−1, 3]
= [12, 11, 14]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 9 / 122
Similarly,
L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 10 / 122
Similarly,
L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])
L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 10 / 122
Similarly,
L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])
L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]
L([x, y, z]) = [−2x + 3y, x − 2y − z, y + 3z]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 10 / 122
Similarly,
L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])
L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]
L([x, y, z]) = [−2x + 3y, x − 2y − z, y + 3z]
Note that
L
xyz
=
−2 3 01 −2 −10 1 3
xyz
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 10 / 122
Exercise: Suppose L : R2 → R
2 is a linear operatorand L([1, 1]) = [3, 0] and L([−1, 1]) = [0, 1]. FindL([x, y]) for all [x, y] ∈ R
2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 11 / 122
Exercise: Suppose L : R2 → R
2 is a linear operatorand L([1, 1]) = [3, 0] and L([−1, 1]) = [0, 1]. FindL([x, y]) for all [x, y] ∈ R
2.
Answer: L([x, y]) =[
3x+3y2
, −x+y2
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 11 / 122
Exercise: Suppose L : R2 → R
2 is a linear operatorand L([1, 1]) = [3, 0] and L([−1, 1]) = [0, 1]. FindL([x, y]) for all [x, y] ∈ R
2.
Answer: L([x, y]) =[
3x+3y2
, −x+y2
]
.
Remark: Let V and W be vector spaces, and letL : V → W be a LT. Also, let {v1, v2, . . . , vn} be abasis for V.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 11 / 122
Exercise: Suppose L : R2 → R
2 is a linear operatorand L([1, 1]) = [3, 0] and L([−1, 1]) = [0, 1]. FindL([x, y]) for all [x, y] ∈ R
2.
Answer: L([x, y]) =[
3x+3y2
, −x+y2
]
.
Remark: Let V and W be vector spaces, and letL : V → W be a LT. Also, let {v1, v2, . . . , vn} be abasis for V. If v ∈ V, L(v) is completely determinedby {L(v1), L(v2), . . . , L(vn)}.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 11 / 122
Composition of Linear transformations
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 12 / 122
Composition of Linear transformations
Theorem 2: Let V1, V2 and V3 be vector spaces andlet L1 : V1 → V2 and L2 : V2 → V3 be lineartransformations. Then L2 ◦ L1 : V1 → V3 given by(L2 ◦ L1)(v) = L2(L1(v)), for all v ∈ V1, is a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 12 / 122
Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 13 / 122
Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.Compute L2 ◦ L1 and L1 ◦ L2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 13 / 122
Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.Compute L2 ◦ L1 and L1 ◦ L2.
Answer:L2 ◦ L1(ax2 + bx + c) = 2ax.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 13 / 122
Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.Compute L2 ◦ L1 and L1 ◦ L2.
Answer:L2 ◦ L1(ax2 + bx + c) = 2ax.
L1 ◦ L2(ax2 + bx + c) = 4ax + b.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 13 / 122
Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.Compute L2 ◦ L1 and L1 ◦ L2.
Answer:L2 ◦ L1(ax2 + bx + c) = 2ax.
L1 ◦ L2(ax2 + bx + c) = 4ax + b.
Clearly , L2 ◦ L1 6= L1 ◦ L2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 13 / 122
Section 5.3: The Dimension Theorem
Kernel of a linear transformation: Let L : V → Wbe a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 14 / 122
Section 5.3: The Dimension Theorem
Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L),
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 14 / 122
Section 5.3: The Dimension Theorem
Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L), is thesubset of all vectors in V that map to 0W , i.e.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 14 / 122
Section 5.3: The Dimension Theorem
Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L), is thesubset of all vectors in V that map to 0W , i.e.
ker(L) = {v ∈ V | L(v) = 0W}.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 14 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}
= {[x, y, z] ∈ R3 | y = 0}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}
= {[x, y, z] ∈ R3 | y = 0}
= {[x, 0, z] | x, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}
= {[x, y, z] ∈ R3 | y = 0}
= {[x, 0, z] | x, z ∈ R}
In this Example, Note that
ker(L) = {[x, 0, z] | x, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Example 6: Let L : R3 → R
2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).
Solution:
ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}
= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}
= {[x, y, z] ∈ R3 | y = 0}
= {[x, 0, z] | x, z ∈ R}
In this Example, Note that
ker(L) = {[x, 0, z] | x, z ∈ R}
is a subspace of the vector space R3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 15 / 122
Result: If L : V → W is a LT, then ker(L) is asubspace of V.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 16 / 122
Range of a linear transformation:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 17 / 122
Range of a linear transformation:
Definition: Let L : V → W be a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 17 / 122
Range of a linear transformation:
Definition: Let L : V → W be a LT. The range of L,denoted by range(L), is the subset of all vectors inW that are image of some vector in V, i.e.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 17 / 122
Range of a linear transformation:
Definition: Let L : V → W be a LT. The range of L,denoted by range(L), is the subset of all vectors inW that are image of some vector in V, i.e.
range(L) = {L(v) | v ∈ V}
Thus a vector w ∈ range(L)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 17 / 122
Range of a linear transformation:
Definition: Let L : V → W be a LT. The range of L,denoted by range(L), is the subset of all vectors inW that are image of some vector in V, i.e.
range(L) = {L(v) | v ∈ V}
Thus a vector w ∈ range(L) implies there existssome vector v ∈ V such that L(v) = w.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 17 / 122
Result: If L : V → W is a LT, then range(L) is asubspace of W.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 18 / 122
Result: If L : V → W is a LT, then range(L) is asubspace of W.
Exercise: Let L : R3 → R
3 be a LT given by
L
xyz
=
5 1 −1−3 0 11 −1 −1
xyz
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 18 / 122
Result: If L : V → W is a LT, then range(L) is asubspace of W.
Exercise: Let L : R3 → R
3 be a LT given by
L
xyz
=
5 1 −1−3 0 11 −1 −1
xyz
Is [1,−2, 3]T ∈ ker(L)?
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 18 / 122
Result: If L : V → W is a LT, then range(L) is asubspace of W.
Exercise: Let L : R3 → R
3 be a LT given by
L
xyz
=
5 1 −1−3 0 11 −1 −1
xyz
Is [1,−2, 3]T ∈ ker(L)?Is [2,−1, 4]T ∈ range(L)?
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 18 / 122
Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 19 / 122
Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T implies[1,−2, 3]T ∈ ker(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 19 / 122
Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T implies[1,−2, 3]T ∈ ker(L).
Note that to check [2,−1, 4]T ∈ range(L) is same asto check whether given system of linear equations
5x + y − z = 2
−3x + z = −1
x − y − z = 4
is consistent or not.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 19 / 122
Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T implies[1,−2, 3]T ∈ ker(L).
Note that to check [2,−1, 4]T ∈ range(L) is same asto check whether given system of linear equations
5x + y − z = 2
−3x + z = −1
x − y − z = 4
is consistent or not.
Since above system of equations is inconsistent(show it!),
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 19 / 122
Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T implies[1,−2, 3]T ∈ ker(L).
Note that to check [2,−1, 4]T ∈ range(L) is same asto check whether given system of linear equations
5x + y − z = 2
−3x + z = −1
x − y − z = 4
is consistent or not.
Since above system of equations is inconsistent(show it!), [2,−1, 4]T /∈ range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 19 / 122
Example 7: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 20 / 122
Example 7: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3
Find range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 20 / 122
Example 7: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3
Find range(L).Find the dimension of ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 20 / 122
Example 7: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3
Find range(L).Find the dimension of ker(L) and range(L).
Solution:
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 20 / 122
Example 7: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3
Find range(L).Find the dimension of ker(L) and range(L).
Solution:
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[0, y] | y ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 20 / 122
range(L) = {y[0, 1] | y ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)
= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)
= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}
= span{[1, 0, 0], [0, 0, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)
= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}
= span{[1, 0, 0], [0, 0, 1]}
Now, the set {[1, 0, 0], [0, 0, 1]} of vectors is LI subsetof R
3 (verify!).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
range(L) = {y[0, 1] | y ∈ R}
Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R
2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1
ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)
= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}
= span{[1, 0, 0], [0, 0, 1]}
Now, the set {[1, 0, 0], [0, 0, 1]} of vectors is LI subsetof R
3 (verify!). Hence, the set {[1, 0, 0], [0, 0, 1]} formsa basis of ker(L) and dim(ker(L)) = 2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 21 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Solution:
ker(L) ={
[x, y, z] ∈ R3 | L([x, y, z]) = 0R2
}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Solution:
ker(L) ={
[x, y, z] ∈ R3 | L([x, y, z]) = 0R2
}
={
[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]
}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Solution:
ker(L) ={
[x, y, z] ∈ R3 | L([x, y, z]) = 0R2
}
={
[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]
}
={
[x, y, z] ∈ R3 | x = 2y, z = −y
}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Solution:
ker(L) ={
[x, y, z] ∈ R3 | L([x, y, z]) = 0R2
}
={
[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]
}
={
[x, y, z] ∈ R3 | x = 2y, z = −y
}
= {[2y, y,−y] | y ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Example 8: Let L : R3 → R
2 be a LT given by
L([x, y, z]) = [x − 2y, y + z].
Find ker(L) and range(L). Also, find basis for ker(L)and range(L).
Solution:
ker(L) ={
[x, y, z] ∈ R3 | L([x, y, z]) = 0R2
}
={
[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]
}
={
[x, y, z] ∈ R3 | x = 2y, z = −y
}
= {[2y, y,−y] | y ∈ R}
= span{[2, 1,−1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 22 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[x − 2y, y + z] | x, y, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[x − 2y, y + z] | x, y, z ∈ R}
= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[x − 2y, y + z] | x, y, z ∈ R}
= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}
= span{[1, 0], [−2, 1], [0, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[x − 2y, y + z] | x, y, z ∈ R}
= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}
= span{[1, 0], [−2, 1], [0, 1]}
= span{[1, 0], [0, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now
range(L) ={
L([x, y, z]) | [x, y, z] ∈ R3}
= {[x − 2y, y + z] | x, y, z ∈ R}
= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}
= span{[1, 0], [−2, 1], [0, 1]}
= span{[1, 0], [0, 1]}
Since the set {[1, 0], [0, 1]} is LI. Thus,
{[1, 0], [0, 1]}
is a basis for range(L).Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122
Exercise: Given a map
L : P3 → P2 given by
L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.
1 Show that L is a linear transformation.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 24 / 122
Exercise: Given a map
L : P3 → P2 given by
L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.
1 Show that L is a linear transformation.2 Find ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 24 / 122
Exercise: Given a map
L : P3 → P2 given by
L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.
1 Show that L is a linear transformation.2 Find ker(L) and range(L).
Answer:
ker(L) ={
0x3 + 0x2 + 0x + d | d ∈ R}
range(L) = P2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 24 / 122
Exercise: Given a map
L : R4 → P2 given by
L([a, b, c, d]) = a + (b + c)x + (b − d)x2.
1 Find ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 25 / 122
Exercise: Given a map
L : R4 → P2 given by
L([a, b, c, d]) = a + (b + c)x + (b − d)x2.
1 Find ker(L) and range(L).2 Find a basis for ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 25 / 122
Exercise: Given a map
L : R4 → P2 given by
L([a, b, c, d]) = a + (b + c)x + (b − d)x2.
1 Find ker(L) and range(L).2 Find a basis for ker(L) and range(L).
Answer:
ker(L) = {[0, b,−b, b] | b ∈ R} and B = {[0, 1,−1, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 25 / 122
Exercise: Given a map
L : R4 → P2 given by
L([a, b, c, d]) = a + (b + c)x + (b − d)x2.
1 Find ker(L) and range(L).2 Find a basis for ker(L) and range(L).
Answer:
ker(L) = {[0, b,−b, b] | b ∈ R} and B = {[0, 1,−1, 1]}
range(L) = span{1, x + x2, x, x2} and B = {1, x, x2}.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 25 / 122
Example 9: Let L : R3 → R
4 be a LT given by
L([x, y, z]) = [x, y − z, x − y + z, x + y − z].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 26 / 122
Example 9: Let L : R3 → R
4 be a LT given by
L([x, y, z]) = [x, y − z, x − y + z, x + y − z].
Find a basis for ker(L) and range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 26 / 122
Example 9: Let L : R3 → R
4 be a LT given by
L([x, y, z]) = [x, y − z, x − y + z, x + y − z].
Find a basis for ker(L) and range(L).
Answer:
{[0, 1, 1]} is a basis of ker(L).{[1, 0, 1, 1], [0, 1,−1, 1]} is a basis for range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 26 / 122
Alternative approach for finding a basis for ker(L)(Kernel Method) Let L : R
n → Rm be a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 27 / 122
Alternative approach for finding a basis for ker(L)(Kernel Method) Let L : R
n → Rm be a LT.
Step 1: Express L(X) = AX for some m × n matrixA.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 27 / 122
Alternative approach for finding a basis for ker(L)(Kernel Method) Let L : R
n → Rm be a LT.
Step 1: Express L(X) = AX for some m × n matrixA. In Example 9, note that L(X) = AX where
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 27 / 122
Alternative approach for finding a basis for ker(L)(Kernel Method) Let L : R
n → Rm be a LT.
Step 1: Express L(X) = AX for some m × n matrixA. In Example 9, note that L(X) = AX where
X =
xyz
and A =
1 0 00 1 −11 −1 11 1 −1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 27 / 122
Step 2: Find matrix B, the RREF of A.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
Step 2: Find matrix B, the RREF of A.
B = RREF(A) =
1 0 00 1 −10 0 00 0 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
Step 2: Find matrix B, the RREF of A.
B = RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 3: Solve the system BX = 0 to find ker(L)such that ker(L) = span(S) for some S ⊆ R
n.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
Step 2: Find matrix B, the RREF of A.
B = RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 3: Solve the system BX = 0 to find ker(L)such that ker(L) = span(S) for some S ⊆ R
n. Thesystem corresponding to B is x = 0, y = z.
ker(L) = {X ∈ Rn|L(X) = AX = 0}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
Step 2: Find matrix B, the RREF of A.
B = RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 3: Solve the system BX = 0 to find ker(L)such that ker(L) = span(S) for some S ⊆ R
n. Thesystem corresponding to B is x = 0, y = z.
ker(L) = {X ∈ Rn|L(X) = AX = 0}
= {X ∈ Rn | BX = 0}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
Step 2: Find matrix B, the RREF of A.
B = RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 3: Solve the system BX = 0 to find ker(L)such that ker(L) = span(S) for some S ⊆ R
n. Thesystem corresponding to B is x = 0, y = z.
ker(L) = {X ∈ Rn|L(X) = AX = 0}
= {X ∈ Rn | BX = 0}
= {[0, y, y] | y ∈ R}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 28 / 122
ker(L) = span{[0, 1, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 29 / 122
ker(L) = span{[0, 1, 1]}
ker(L) = span(S), where S = {[0, 1, 1]}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 29 / 122
ker(L) = span{[0, 1, 1]}
ker(L) = span(S), where S = {[0, 1, 1]}
Step 4: Find a LI subset of S which forms a basis forker(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 29 / 122
ker(L) = span{[0, 1, 1]}
ker(L) = span(S), where S = {[0, 1, 1]}
Step 4: Find a LI subset of S which forms a basis forker(L). Since the set {[0, 1, 1]} is a LI so it is a basisof ker(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 29 / 122
Alternative approach to find a basis for range (L)(Range Method)
Step 1: Find RREF of A.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 30 / 122
Alternative approach to find a basis for range (L)(Range Method)
Step 1: Find RREF of A.
RREF(A) =
1 0 00 1 −10 0 00 0 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 30 / 122
Alternative approach to find a basis for range (L)(Range Method)
Step 1: Find RREF of A.
RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 30 / 122
Alternative approach to find a basis for range (L)(Range Method)
Step 1: Find RREF of A.
RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).Note that, Columns I and II have leading entry.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 30 / 122
Alternative approach to find a basis for range (L)(Range Method)
Step 1: Find RREF of A.
RREF(A) =
1 0 00 1 −10 0 00 0 0
Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).Note that, Columns I and II have leading entry. Thus,the corresponding column vector of A i.e.{[1, 0, 1, 1], [0, 1,−1, 1]} is a basis of range (L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 30 / 122
The Dimension Theorem:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 31 / 122
The Dimension Theorem: If L : V → W is a LT andV is finite dimensional, then range(L) is finitedimensional, and
dim(ker(L)) + dim(range(L)) = dim(V).
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The Dimension Theorem: If L : V → W is a LT andV is finite dimensional, then range(L) is finitedimensional, and
dim(ker(L)) + dim(range(L)) = dim(V).
Sometimes dim(ker(L)) and dim(range(L)) is alsoknown as nullity (L) and rank (L), respectively.
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Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
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Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 32 / 122
Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Solution:
ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}
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Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Solution:
ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}
ker(L) = {0P2}
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 32 / 122
Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Solution:
ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}
ker(L) = {0P2} implies dim(ker(L)) = 0.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 32 / 122
Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Solution:
ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}
ker(L) = {0P2} implies dim(ker(L)) = 0.
Since dimP2 = 3 by dimension theorem, we have
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 32 / 122
Example 10: Consider a LT L : P2 → P3 given by
L(a + bx + cx2) = x(a + bx + cx2).
Find dim(ker(L)) and dim(range(L)).
Solution:
ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}
ker(L) = {0P2} implies dim(ker(L)) = 0.
Since dimP2 = 3 by dimension theorem, we have
dim(range(L)) = 3 − 0 = 3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 32 / 122
Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)
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Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
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Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
Find ker(L), dim(ker(L)), range(L) anddim(range(L)).
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Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
Find ker(L), dim(ker(L)), range(L) anddim(range(L)).
Answer:
ker(L) =
a b cd e fg h −a − e
3×3
| a, b, c, d, e, f, g, h ∈ R
Note that
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 33 / 122
Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
Find ker(L), dim(ker(L)), range(L) anddim(range(L)).
Answer:
ker(L) =
a b cd e fg h −a − e
3×3
| a, b, c, d, e, f, g, h ∈ R
Note that dim(ker(L)) = 8 (show it).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 33 / 122
Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
Find ker(L), dim(ker(L)), range(L) anddim(range(L)).
Answer:
ker(L) =
a b cd e fg h −a − e
3×3
| a, b, c, d, e, f, g, h ∈ R
Note that dim(ker(L)) = 8 (show it). Sincerange(L) = R
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 33 / 122
Exercise: Consider a LT L : M33 → R given by
L(A) = trace(A)(sum of the diagonal entries ofA).
Find ker(L), dim(ker(L)), range(L) anddim(range(L)).
Answer:
ker(L) =
a b cd e fg h −a − e
3×3
| a, b, c, d, e, f, g, h ∈ R
Note that dim(ker(L)) = 8 (show it). Sincerange(L) = R so that dim(range(L)) = 1.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 33 / 122
Exercise: Let W be the vector space of all 2 × 2symmetric matrices. Define a LT L : W → P2 by
L
([
a bb c
])
= (a − b) + (b − c)x + (c − a)x2
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Exercise: Let W be the vector space of all 2 × 2symmetric matrices. Define a LT L : W → P2 by
L
([
a bb c
])
= (a − b) + (b − c)x + (c − a)x2
Find dim(ker(L)) and dim(range(L)).
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Exercise: Let W be the vector space of all 2 × 2symmetric matrices. Define a LT L : W → P2 by
L
([
a bb c
])
= (a − b) + (b − c)x + (c − a)x2
Find dim(ker(L)) and dim(range(L)).
Answer: dim(ker(L)) = 1 and dim(range(L)) = 2.
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Exercise: Let {e1, e2, e3, e4} be standard basis for R4
and L : R4 → R
3 be a LT given by
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Exercise: Let {e1, e2, e3, e4} be standard basis for R4
and L : R4 → R
3 be a LT given by
L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]
L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]
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Exercise: Let {e1, e2, e3, e4} be standard basis for R4
and L : R4 → R
3 be a LT given by
L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]
L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]
Find ker(L) and dim(ker(L)).Find range(L) and dim(range(L)).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 35 / 122
Exercise: Let {e1, e2, e3, e4} be standard basis for R4
and L : R4 → R
3 be a LT given by
L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]
L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]
Find ker(L) and dim(ker(L)).Find range(L) and dim(range(L)).
Answer: dim(ker(L)) = 1 and dim(range(L)) = 3.
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Exercise: For each p ∈ P2, consider L : P2 → P4
given by L(p) = x2p.
Find ker(L) and dim(ker(L)).
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Exercise: For each p ∈ P2, consider L : P2 → P4
given by L(p) = x2p.
Find ker(L) and dim(ker(L)).Find range(L) and dim(range(L)).
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Exercise: For each p ∈ P2, consider L : P2 → P4
given by L(p) = x2p.
Find ker(L) and dim(ker(L)).Find range(L) and dim(range(L)).
Answer: dim(ker(L)) = 0 and dim(range(L)) = 3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 36 / 122
Section 5.4
Definition: A linear transformation L : V → Wone-to-one if and only if
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Section 5.4
Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2,
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Section 5.4
Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2, or if v1 6= v2 impliesL(v1) 6= L(v2).
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Section 5.4
Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2, or if v1 6= v2 impliesL(v1) 6= L(v2).
L is onto if and only if,
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Section 5.4
Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2, or if v1 6= v2 impliesL(v1) 6= L(v2).
L is onto if and only if, for each w ∈ W, there issome v ∈ V such that L(v) = w.
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Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
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Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 38 / 122
Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.
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Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 38 / 122
Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies L is not one-to-one.
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Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies L is not one-to-one.
Let q be an arbitrary element in P2 i.e.q = a + bx + cx2.
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Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies L is not one-to-one.
Let q be an arbitrary element in P2 i.e.q = a + bx + cx2. Note that a + bx + cx2 = p′, wherep = ax +
(
b2
)
x2 +(
c3
)
x3 so that L(p) = q.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 38 / 122
Example 11: Consider a LT
L : P3 → P2 given by
L(p) = p′.
Check if L is one-to-one and onto.
Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies L is not one-to-one.
Let q be an arbitrary element in P2 i.e.q = a + bx + cx2. Note that a + bx + cx2 = p′, wherep = ax +
(
b2
)
x2 +(
c3
)
x3 so that L(p) = q. Hence, Lis onto.
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Exercise: Which of the following transformations areone-to-one? onto?
1 L : R2 → R
3 given by L([x, y]) = [2x, x− y, 0].2 L : R
3 → R4 given by L([x, y, z]) = [y, z,−y, 0].
3 L : M22 → M22 given by L(A) = AT .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 39 / 122
Exercise: Which of the following transformations areone-to-one? onto?
1 L : R2 → R
3 given by L([x, y]) = [2x, x− y, 0].2 L : R
3 → R4 given by L([x, y, z]) = [y, z,−y, 0].
3 L : M22 → M22 given by L(A) = AT .
Answer:1 one-to-one but not onto.
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Exercise: Which of the following transformations areone-to-one? onto?
1 L : R2 → R
3 given by L([x, y]) = [2x, x− y, 0].2 L : R
3 → R4 given by L([x, y, z]) = [y, z,−y, 0].
3 L : M22 → M22 given by L(A) = AT .
Answer:1 one-to-one but not onto.2 neither one-to-one nor onto
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 39 / 122
Exercise: Which of the following transformations areone-to-one? onto?
1 L : R2 → R
3 given by L([x, y]) = [2x, x− y, 0].2 L : R
3 → R4 given by L([x, y, z]) = [y, z,−y, 0].
3 L : M22 → M22 given by L(A) = AT .
Answer:1 one-to-one but not onto.2 neither one-to-one nor onto3 one-to-one and onto.
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Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 40 / 122
Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then L is one-to-one if and onlyif ker(L) = {0V} (i.e., dimker(L) = 0).
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Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then L is one-to-one if and onlyif ker(L) = {0V} (i.e., dimker(L) = 0).
Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT.
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Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then L is one-to-one if and onlyif ker(L) = {0V} (i.e., dimker(L) = 0).
Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT. If W is finite dimensional, then Lis onto
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 40 / 122
Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then L is one-to-one if and onlyif ker(L) = {0V} (i.e., dimker(L) = 0).
Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT. If W is finite dimensional, then Lis onto if and only if dim(range(L)) = dim(W).
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Example 12: Consider a LT L : M22 → M23 givenby
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
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Example 12: Consider a LT L : M22 → M23 givenby
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
Is L one-to-one and onto?
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 41 / 122
Example 12: Consider a LT L : M22 → M23 givenby
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
Is L one-to-one and onto?
Solution: Let[
a bc d
]
∈ ker(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 41 / 122
Example 12: Consider a LT L : M22 → M23 givenby
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
Is L one-to-one and onto?
Solution: Let[
a bc d
]
∈ ker(L). Then
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
=
[
0 0 00 0 0
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 41 / 122
Example 12: Consider a LT L : M22 → M23 givenby
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
Is L one-to-one and onto?
Solution: Let[
a bc d
]
∈ ker(L). Then
L
([
a bc d
])
=
[
a − b 0 c − dc + d a + b 0
]
=
[
0 0 00 0 0
]
.
We have a − b = c − d = c + d = a + b = 0 impliesa = b = c = d = 0.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 41 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.
Note that
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.
Note that
dim(range(L)) = dim(M22) − dim(ker(L))
= 4
6= dim(M23).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.
Note that
dim(range(L)) = dim(M22) − dim(ker(L))
= 4
6= dim(M23).
Hence, L is not onto.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.
Note that
dim(range(L)) = dim(M22) − dim(ker(L))
= 4
6= dim(M23).
Hence, L is not onto.
Try to find a basis of range(L).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 42 / 122
Example 13: Consider a LT L : R3 → R
3 given by
L
xyz
=
−7 4 −216 −7 24 −3 2
xyz
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 43 / 122
Example 13: Consider a LT L : R3 → R
3 given by
L
xyz
=
−7 4 −216 −7 24 −3 2
xyz
Is L one-to-one and onto?
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 43 / 122
Example 13: Consider a LT L : R3 → R
3 given by
L
xyz
=
−7 4 −216 −7 24 −3 2
xyz
Is L one-to-one and onto?
Solution: The RREF of matrix A =
−7 4 −216 −7 24 −3 2
is
1 0 −2
5
0 1 −6
5
0 0 0
.
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From range method,
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 44 / 122
From range method, dim(range(L)) = 2
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 44 / 122
From range method, dim(range(L)) = 2 and byDimension theorem, dim(ker(L)) = 1.
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From range method, dim(range(L)) = 2 and byDimension theorem, dim(ker(L)) = 1. Hence, L isneither one-to-one nor onto.
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA.
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) =
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n.
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L)
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 45 / 122
Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one. ByDimension theorem, we see that
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one. ByDimension theorem, we see that
dim(range(L)) = n2 − dim(ker(L))
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one. ByDimension theorem, we see that
dim(range(L)) = n2 − dim(ker(L))
6= n2
6= dimMnn
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Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?
Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one. ByDimension theorem, we see that
dim(range(L)) = n2 − dim(ker(L))
6= n2
6= dimMnn
Hence, L is not onto.
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x).
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
implies ker(L) = {0P}.
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
implies ker(L) = {0P}. Hence, L is one-to-one.
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L),
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L), L is not onto.
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Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?
Solution:
ker(L) = {p(x)|L(p(x)) = 0P}
implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L), L is not onto.
Question : Can we apply Dimension theorem here?
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
Is L one-to-one and onto?
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
Is L one-to-one and onto?
Answer: L is onto but not one-to-one.
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
Is L one-to-one and onto?
Answer: L is onto but not one-to-one.
Exercise: Consider a LT L : P2 → P2 given by
L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c).
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
Is L one-to-one and onto?
Answer: L is onto but not one-to-one.
Exercise: Consider a LT L : P2 → P2 given by
L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c). Is Lone-to-one and onto?
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Exercise: Consider a LT L : M23 → M22 given by
L
([
a b cd e f
])
=
[
a + b a + cd + e d + f
]
Is L one-to-one and onto?
Answer: L is onto but not one-to-one.
Exercise: Consider a LT L : P2 → P2 given by
L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c). Is Lone-to-one and onto?
Answer: L is one-to-one and onto.
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Exercise: Consider a LT L : R4 → R
3 given by
L
x1
x2
x3
x4
=
−5 3 1 18−2 1 1 6−7 3 4 19
x1
x2
x3
x4
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Exercise: Consider a LT L : R4 → R
3 given by
L
x1
x2
x3
x4
=
−5 3 1 18−2 1 1 6−7 3 4 19
x1
x2
x3
x4
Is L one-to-one and onto?
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Exercise: Consider a LT L : R4 → R
3 given by
L
x1
x2
x3
x4
=
−5 3 1 18−2 1 1 6−7 3 4 19
x1
x2
x3
x4
Is L one-to-one and onto?
Answer: L is not one-to-one but onto.
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Exercise: Consider a LT L : R4 → R
3 given by
L
x1
x2
x3
x4
=
−5 3 1 18−2 1 1 6−7 3 4 19
x1
x2
x3
x4
Is L one-to-one and onto?
Answer: L is not one-to-one but onto.
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Section 5.5
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Section 5.5
Invertible linear transformation:
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Section 5.5
Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if
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Section 5.5
Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if there isa function M : W → V such that (M ◦ L)(v) = v, forall v ∈ V, and (L ◦ M)(w) = w, for all w ∈ W.
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Section 5.5
Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if there isa function M : W → V such that (M ◦ L)(v) = v, forall v ∈ V, and (L ◦ M)(w) = w, for all w ∈ W.
Such a function M , denoted by L−1, is called aninverse of L.
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.
Solution: First we need to show that L is a linearoperator.
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.
Solution: First we need to show that L is a linearoperator.
L(p + q) = (p + q) + (p + q)′
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.
Solution: First we need to show that L is a linearoperator.
L(p + q) = (p + q) + (p + q)′
= p + p′ + q + q′
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Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.
Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.
Solution: First we need to show that L is a linearoperator.
L(p + q) = (p + q) + (p + q)′
= p + p′ + q + q′
= L(p) + L(q) for all p, q ∈ Pn.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
= {0Pn}
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
= {0Pn}
implies L is one-to-one.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
= {0Pn}
implies L is one-to-one.
By Dimension theorem,
dim range(L) = dimPn = n + 1
so that range(L) = Pn.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
= {0Pn}
implies L is one-to-one.
By Dimension theorem,
dim range(L) = dimPn = n + 1
so that range(L) = Pn. Thus, L is onto.
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Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.
kerL = {p ∈ Pn | L(p) = 0Pn}
= {p ∈ Pn | p + p′ = 0Pn}
= {0Pn}
implies L is one-to-one.
By Dimension theorem,
dim range(L) = dimPn = n + 1
so that range(L) = Pn. Thus, L is onto. Hence, L isan isomorphism.
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3.
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3. Note
thatL([x, y, z]) = [x + z, x + y + z, y + z].
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3. Note
thatL([x, y, z]) = [x + z, x + y + z, y + z].
and ker(L) = {0R3}.
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3. Note
thatL([x, y, z]) = [x + z, x + y + z, y + z].
and ker(L) = {0R3}. Thus, L is one-to-one.
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Example 17: Show that the linear operator
L : R3 → R
3 such that
L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3
is an isomorphism.
Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3. Note
thatL([x, y, z]) = [x + z, x + y + z, y + z].
and ker(L) = {0R3}. Thus, L is one-to-one. Usedimension theorem and Theorem 4 to conclude L isonto. Hence, L is an isomorphism.
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Result: Let L : V → W be a linear transformation,where V and W be finite dimensional vector spacessuch that dim(V) = dim(W). Then L is one-to-one ifand only if L is onto.
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Exercise: Show that the linear operator L : P2 → P2
given by L(a+ bx+ cx2) = (b+ c)+(a+ c)x+(a+ b)x2
is an isomorphism.
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Exercise: Show that the linear operator L : P2 → P2
given by L(a+ bx+ cx2) = (b+ c)+(a+ c)x+(a+ b)x2
is an isomorphism.
Exercise: Show that the linear transformationL : Mmn → Mnm given by L(A) = AT is anisomorphism.
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT.
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.
Example 18: Let L : R3 → P2 be a LT given by
L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.
Example 18: Let L : R3 → P2 be a LT given by
L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.
Is L invertible?
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.
Example 18: Let L : R3 → P2 be a LT given by
L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.
Is L invertible? If yes, find L−1.
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Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.
Example 18: Let L : R3 → P2 be a LT given by
L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.
Is L invertible? If yes, find L−1.
Solution: First show that L is both one-to-one andonto. Hence, invertible.
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Let L−1 : P2 → R3 be defined by
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 56 / 122
Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 56 / 122
Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
⇒ L([x, y, z]) = a + bt + ct2
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Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
⇒ L([x, y, z]) = a + bt + ct2
⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2
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Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
⇒ L([x, y, z]) = a + bt + ct2
⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2
⇒ x = a, x + y − z = b, x + y + z = c
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Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
⇒ L([x, y, z]) = a + bt + ct2
⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2
⇒ x = a, x + y − z = b, x + y + z = c
⇒ x = a, y =b + c − 2a
2, z =
c − b
2.
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Let L−1 : P2 → R3 be defined by
L−1(a + bt + ct2) = [x, y, z]
⇒ L([x, y, z]) = a + bt + ct2
⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2
⇒ x = a, x + y − z = b, x + y + z = c
⇒ x = a, y =b + c − 2a
2, z =
c − b
2.
Hence, L−1(a + bx + cx2) =[
a, b+c−2a2
, c−b2
]
.
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Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2.
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Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible?
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Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible? If yes, find L−1.
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Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible? If yes, find L−1.
Answer:
L−1(a+bx+cx2) =1
2(b+c−a)+
1
2(a+c−b)x+
1
2(a+b−c)x2.
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Exercise: Let L : R3 → R
3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.
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Exercise: Let L : R3 → R
3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible?
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Exercise: Let L : R3 → R
3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible? If yes, find L−1.
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Exercise: Let L : R3 → R
3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible? If yes, find L−1.
Answer: L−1([x, y, z]) = [y − z, y − x, x − y + z].
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Isomorphic vector spaces: Let V and W be vectorspaces. Then V is isomorphic to W, denoted byV ∼= W, if and only if there exists an isomorphismL : V → W.
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Isomorphic vector spaces: Let V and W be vectorspaces. Then V is isomorphic to W, denoted byV ∼= W, if and only if there exists an isomorphismL : V → W.
Theorem 5: Suppose V ∼= W and V and W are finitedimensional. Then V is isomorphic to W if and only ifdim(V) = dim(W).
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Exercise: Show that Rn and Pn are not isomorphic.
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Exercise: Show that Rn and Pn are not isomorphic.
Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 60 / 122
Exercise: Show that Rn and Pn are not isomorphic.
Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),R
n and Pn are not isomorphic.
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Exercise: Show that Rn and Pn are not isomorphic.
Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),R
n and Pn are not isomorphic.
Exercise: Let W be the vector space of allsymmetric 2 × 2 matrices. Show that W isisomorphic to R
3.
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Exercise: Show that the subspace
W = {p ∈ P3 | p(0) = 0}
is isomorphic to P2.
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Section 4.7
Ordered Basis: An ordered basis for vector spaceV is an ordered n-tuple of vectors (v1, v2, . . . , vn)such that the set {v1, v2, . . . , vn} is a basis for V.
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Section 4.7
Ordered Basis: An ordered basis for vector spaceV is an ordered n-tuple of vectors (v1, v2, . . . , vn)such that the set {v1, v2, . . . , vn} is a basis for V.
(e1, e2) and (e2, e1) are two ordered bases for R2.
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Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V.
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Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V. Suppose thatw ∈ V such that
w = a1v1 + · · · + anvn
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Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V. Suppose thatw ∈ V such that
w = a1v1 + · · · + anvn
Then [w]B, the coordinatization or coordinates of wwith respect to B is the n-vector [a1, a2, . . . , an].
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Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R
2.
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Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R
2. Note that
[4, 2] = 1[4, 2] + 0[1, 3].
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Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R
2. Note that
[4, 2] = 1[4, 2] + 0[1, 3].
Hence, [4, 2]B = [1, 0].
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Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R
2. Note that
[4, 2] = 1[4, 2] + 0[1, 3].
Hence, [4, 2]B = [1, 0]. Similarly,
[11, 13] = 2[4, 2] + 3[1, 3].
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Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R
2. Note that
[4, 2] = 1[4, 2] + 0[1, 3].
Hence, [4, 2]B = [1, 0]. Similarly,
[11, 13] = 2[4, 2] + 3[1, 3].
Hence, [11, 13]B = [2, 3].
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Example 20: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace V of R5.
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Example 20: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace V of R5.
Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.
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Example 20: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace V of R5.
Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.
Solution: To find [−23, 30,−7,−1,−7]B,
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 65 / 122
Example 20: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace V of R5.
Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.
Solution: To find [−23, 30,−7,−1,−7]B, we need tosolve the following equation[−23, 30,−7,−1,−7] =a[−4, 5,−1, 0,−1] + b[1,−3, 2, 2, 5] + c[1,−2, 1, 1, 3]
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or equivalently
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or equivalently
−4a + b + c = −23
5a − 3b − 2c = 30
−a + 2b + c = −7
2b + c = −1
−a + 5b + 3c = −7
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or equivalently
−4a + b + c = −23
5a − 3b − 2c = 30
−a + 2b + c = −7
2b + c = −1
−a + 5b + 3c = −7
To solve this system, note that the RREF of theaugmented matrix
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 66 / 122
−4 1 1 −235 −3 −2 30
−1 2 1 −70 2 1 −1
−1 5 3 −7
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−4 1 1 −235 −3 −2 30
−1 2 1 −70 2 1 −1
−1 5 3 −7
is
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−4 1 1 −235 −3 −2 30
−1 2 1 −70 2 1 −1
−1 5 3 −7
is
1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0
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−4 1 1 −235 −3 −2 30
−1 2 1 −70 2 1 −1
−1 5 3 −7
is
1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0
Hence, the unique solution for the system is
a = 6, b = −2, c = 3
implies
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 67 / 122
−4 1 1 −235 −3 −2 30
−1 2 1 −70 2 1 −1
−1 5 3 −7
is
1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0
Hence, the unique solution for the system is
a = 6, b = −2, c = 3
implies
[−23, 30,−7,−1,−7]B = [6,−2, 3].
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To find [1, 2, 3, 4, 5]B, we need solve the followingsystem
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To find [1, 2, 3, 4, 5]B, we need solve the followingsystem
−4a + b + c = 1
5a − 3b − 2c = 2
−a + 2b + c = 3
2b + c = 4
−a + 5b + 3c = 5
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To find [1, 2, 3, 4, 5]B, we need solve the followingsystem
−4a + b + c = 1
5a − 3b − 2c = 2
−a + 2b + c = 3
2b + c = 4
−a + 5b + 3c = 5
To solve this system, note that the RREF of
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 68 / 122
−4 1 1 15 −3 −2 2
−1 2 1 30 2 1 4
−1 5 3 5
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−4 1 1 15 −3 −2 2
−1 2 1 30 2 1 4
−1 5 3 5
is
1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0
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−4 1 1 15 −3 −2 2
−1 2 1 30 2 1 4
−1 5 3 5
is
1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0
This system has no solution,
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−4 1 1 15 −3 −2 2
−1 2 1 30 2 1 4
−1 5 3 5
is
1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0
This system has no solution, implies that the vector[1, 2, 3, 4, 5] is not in span(B) = V.
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Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n.
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Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order,
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]),
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V,
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V, i.e., coordinatization isnot possible.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 70 / 122
Coordinatization Method:Let V be a nontrivial subspace of R
n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R
n. To compute [v]B, we perform the followingsteps:
Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V, i.e., coordinatization isnot possible. Otherwise, v ∈ span(B) = V.
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Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y].
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Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y]. Then, [v]B = y,
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Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y]. Then, [v]B = y, the lastcolumn of [Ik|y].
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Example 21: Let
B =
([
1 −20 1
]
,
[
2 −11 0
]
,
[
1 −13 1
])
be an ordered basis of the subspace W of M22.
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Example 21: Let
B =
([
1 −20 1
]
,
[
2 −11 0
]
,
[
1 −13 1
])
be an ordered basis of the subspace W of M22.
Compute [v]B if exists, where v =
[
−3 −20 3
]
.
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Solution: Consider
[A|v] =
1 2 1 −3−2 −1 −1 −2
0 1 3 01 0 1 3
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Solution: Consider
[A|v] =
1 2 1 −3−2 −1 −1 −2
0 1 3 01 0 1 3
Note that the row reduced echelon form is
RREF[A|v] =
1 0 0 20 1 0 −30 0 1 10 0 0 0
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The row reduced matrix contains no rows with allzero entries on the left and a nonzero entry on theright, so [v]B exists,
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The row reduced matrix contains no rows with allzero entries on the left and a nonzero entry on theright, so [v]B exists, and
[v]B = [2,−3, 1].
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Fundamental properties of Coordinatization: LetB = (v1, v2, . . . , vk) be an ordered basis for a vectorspace V. Suppose w1, w2, . . . , wk ∈ V anda1, a2, . . . , ak are scalars. Then
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 75 / 122
Fundamental properties of Coordinatization: LetB = (v1, v2, . . . , vk) be an ordered basis for a vectorspace V. Suppose w1, w2, . . . , wk ∈ V anda1, a2, . . . , ak are scalars. Then
[w1 + w2]B = [w1]B + [w2]B
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Fundamental properties of Coordinatization: LetB = (v1, v2, . . . , vk) be an ordered basis for a vectorspace V. Suppose w1, w2, . . . , wk ∈ V anda1, a2, . . . , ak are scalars. Then
[w1 + w2]B = [w1]B + [w2]B[a1w1]B = a1[w1]B
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 75 / 122
Fundamental properties of Coordinatization: LetB = (v1, v2, . . . , vk) be an ordered basis for a vectorspace V. Suppose w1, w2, . . . , wk ∈ V anda1, a2, . . . , ak are scalars. Then
[w1 + w2]B = [w1]B + [w2]B[a1w1]B = a1[w1]B[a1w1 + a2w2 + . . . akwk]B= a1[w1]B + a2[w2]B + · · · + ak[wk]B.
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Exercise: Let
B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)
be an ordered basis of the subspace W of P2.
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Exercise: Let
B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)
be an ordered basis of the subspace W of P2.Compute [v]B if exists, where v = 13x2 − 5x + 20.
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Exercise: Let
B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)
be an ordered basis of the subspace W of P2.Compute [v]B if exists, where v = 13x2 − 5x + 20.
Answer: [v]B = [4,−5, 3].
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Exercise: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace W of R5.
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Exercise: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace W of R5.
Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5].
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Exercise: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace W of R5.
Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5]. Compute [2x − 7y + 3z]B.
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Exercise: Let
B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])
be an ordered basis of the subspace W of R5.
Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5]. Compute [2x − 7y + 3z]B.
Answer: [2x − 7y + 3z]B = [−2, 9,−15].
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Example 22: LetC = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3]) bean ordered basis of the subspace W of R
5.
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Example 22: LetC = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3]) bean ordered basis of the subspace W of R
5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W.
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Example 22: LetC = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3]) bean ordered basis of the subspace W of R
5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W. Also, compute[x]C, [y]C, [z]C.
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Example 22: LetC = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3]) bean ordered basis of the subspace W of R
5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W. Also, compute[x]C, [y]C, [z]C.
Solution: We have the following augmented matrix
[
A x y z]
=
−4 1 1 1 0 05 −3 −2 0 1 0
−1 2 1 −1 −1 00 2 1 0 0 1
−1 5 3 4 3 5
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Row reduce echelon form of the above matrix is
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 79 / 122
Row reduce echelon form of the above matrix is
1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 79 / 122
Row reduce echelon form of the above matrix is
1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0
Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7].
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Row reduce echelon form of the above matrix is
1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0
Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7]. Here, the matrix
P =
1 1 1−5 −4 −310 8 7
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 79 / 122
Row reduce echelon form of the above matrix is
1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0
Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7]. Here, the matrix
P =
1 1 1−5 −4 −310 8 7
is called the transition matrix
from B-coordinates to C-coordinates.Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 79 / 122
Transition Matrix:
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Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C.
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Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B.
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Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B. Then P is called the transition matrixfrom B-coordinates to C-coordinates
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Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B. Then P is called the transition matrixfrom B-coordinates to C-coordinates (or transitionmatrix from B to C).
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Transition Matrix Method:
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Transition Matrix Method: To find the transitionmatrix P from B to C,
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Transition Matrix Method: To find the transitionmatrix P from B to C, we apply row reduction on
1st 2nd kth 1st 2nd kth
vector vector ·· vector vector vector ·· vectorin in in in in inC C C B B B
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 81 / 122
Transition Matrix Method: To find the transitionmatrix P from B to C, we apply row reduction on
1st 2nd kth 1st 2nd kth
vector vector ·· vector vector vector ·· vectorin in in in in inC C C B B B
to produce[
Ik Prows of zeroes
]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 81 / 122
Example 23: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 82 / 122
Example 23: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
and
C =
([
22 70 2
]
,
[
12 40 1
]
,
[
33 120 2
])
of U2 (the set of 2 × 2 upper triangular matrices),
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 82 / 122
Example 23: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
and
C =
([
22 70 2
]
,
[
12 40 1
]
,
[
33 120 2
])
of U2 (the set of 2 × 2 upper triangular matrices), findthe transition matrix P from B to C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 82 / 122
Solution: Apply row reduction on
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 83 / 122
Solution: Apply row reduction on
22 12 33 7 1 17 4 12 3 2 −10 0 0 0 0 02 1 2 0 −1 1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 83 / 122
Solution: Apply row reduction on
22 12 33 7 1 17 4 12 3 2 −10 0 0 0 0 02 1 2 0 −1 1
we get
1 0 0 1 −2 10 1 0 −4 1 10 0 1 1 1 −10 0 0 0 0 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 83 / 122
The transition matrix P from B to C is
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 84 / 122
The transition matrix P from B to C is
1 −2 1−4 1 11 1 −1
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 84 / 122
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 85 / 122
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3. Let
P =
1 1 22 1 1−1 −1 1
be the transition matrix from B to C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 85 / 122
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3. Let
P =
1 1 22 1 1−1 −1 1
be the transition matrix from B to C. Find the basisB.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 85 / 122
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3. Let
P =
1 1 22 1 1−1 −1 1
be the transition matrix from B to C. Find the basisB.
Solution:
x = 1 · a + 2 · b − 1 · c = [3, 2, 0]
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3. Let
P =
1 1 22 1 1−1 −1 1
be the transition matrix from B to C. Find the basisB.
Solution:
x = 1 · a + 2 · b − 1 · c = [3, 2, 0]
Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R
3. Let
P =
1 1 22 1 1−1 −1 1
be the transition matrix from B to C. Find the basisB.
Solution:
x = 1 · a + 2 · b − 1 · c = [3, 2, 0]
y = 1 · a + 1 · b − 1 · c = [2, 1, 0]
z = 2 · a + 1 · b + 1 · c = [3, 1, 3].Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 85 / 122
Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 86 / 122
Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).
Change of Coordinates Using the TransitionMatrix
Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 86 / 122
Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).
Change of Coordinates Using the TransitionMatrix
Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be an n × n matrix.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 86 / 122
Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).
Change of Coordinates Using the TransitionMatrix
Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be an n × n matrix. Then P is thetransition matrix from B to C if and only if for everyv ∈ V, P [v]B = [v]C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 86 / 122
Example 25: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 87 / 122
Example 25: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
and
C =
([
22 70 2
]
,
[
12 40 1
]
,
[
33 120 2
])
of U2 (set of 2 × 2 upper triangular matrices).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 87 / 122
Example 25: For the ordered bases
B =
([
7 30 0
]
,
[
1 20 −1
]
,
[
1 −10 1
])
and
C =
([
22 70 2
]
,
[
12 40 1
]
,
[
33 120 2
])
of U2 (set of 2 × 2 upper triangular matrices). Find
[v]B and [v]C, where v =
[
25 240 −9
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 87 / 122
Solution: Clearly,[
25 240 −9
]
= 4
[
7 30 0
]
+ 3
[
1 20 −1
]
− 6
[
1 −10 1
]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 88 / 122
Solution: Clearly,[
25 240 −9
]
= 4
[
7 30 0
]
+ 3
[
1 20 −1
]
− 6
[
1 −10 1
]
Hence, [v]B = [4, 3,−6]T .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 88 / 122
Solution: Clearly,[
25 240 −9
]
= 4
[
7 30 0
]
+ 3
[
1 20 −1
]
− 6
[
1 −10 1
]
Hence, [v]B = [4, 3,−6]T . Now, since [v]C = P [v]Band
P =
1 −2 1−4 1 11 1 −1
(see Example 23)
implies
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 88 / 122
Solution: Clearly,[
25 240 −9
]
= 4
[
7 30 0
]
+ 3
[
1 20 −1
]
− 6
[
1 −10 1
]
Hence, [v]B = [4, 3,−6]T . Now, since [v]C = P [v]Band
P =
1 −2 1−4 1 11 1 −1
(see Example 23)
implies [v]C = [−8,−19, 13]T .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 88 / 122
Solution: Clearly,[
25 240 −9
]
= 4
[
7 30 0
]
+ 3
[
1 20 −1
]
− 6
[
1 −10 1
]
Hence, [v]B = [4, 3,−6]T . Now, since [v]C = P [v]Band
P =
1 −2 1−4 1 11 1 −1
(see Example 23)
implies [v]C = [−8,−19, 13]T . Clearly,[
25 240 −9
]
= −8
[
22 70 2
]
− 19
[
12 40 1
]
+ 13
[
33 120 2
]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 88 / 122
Theorem: Let B and C be be ordered bases for anontrivial finite dimensional vector space V, and letP be the transition matrix from B to C. Then P isnonsingular, and P−1 is the transition matrix from Cto B.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 89 / 122
Example 26: For an ordered basisB = ([1,−4, 1, 2, 1], [6,−24, 5, 8, 3], [3,−12, 3, 6, 2]) of asubspace V of R
5.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 90 / 122
Example 26: For an ordered basisB = ([1,−4, 1, 2, 1], [6,−24, 5, 8, 3], [3,−12, 3, 6, 2]) of asubspace V of R
5.Use the Simplified Span Method to find asecond ordered basis C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 90 / 122
Example 26: For an ordered basisB = ([1,−4, 1, 2, 1], [6,−24, 5, 8, 3], [3,−12, 3, 6, 2]) of asubspace V of R
5.Use the Simplified Span Method to find asecond ordered basis C.
Solution: Consider
B =
1 −4 1 2 16 −24 5 8 33 −12 3 6 2
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 90 / 122
Example 26: For an ordered basisB = ([1,−4, 1, 2, 1], [6,−24, 5, 8, 3], [3,−12, 3, 6, 2]) of asubspace V of R
5.Use the Simplified Span Method to find asecond ordered basis C.
Solution: Consider
B =
1 −4 1 2 16 −24 5 8 33 −12 3 6 2
Note that
RREF(B) =
1 −4 0 −2 00 0 1 4 00 0 0 0 1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 90 / 122
C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 91 / 122
C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .
Find the transition matrix P from B to C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 91 / 122
C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .
Find the transition matrix P from B to C.
Answer:
P =
1 6 31 5 31 3 2
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 91 / 122
Find the transition matrix Q from C to B.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 92 / 122
Find the transition matrix Q from C to B.
Answer:
Q = P−1 =
1 −3 31 −1 0−2 3 −1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 92 / 122
For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 93 / 122
For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.
[B|v] =
1 6 3 2−4 −24 −12 −8
1 5 3 −22 8 6 −121 3 2 3
=
1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 93 / 122
For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.
[B|v] =
1 6 3 2−4 −24 −12 −8
1 5 3 −22 8 6 −121 3 2 3
=
1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0
Thus,[v]B = [17, 4,−13]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 93 / 122
For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.
[B|v] =
1 6 3 2−4 −24 −12 −8
1 5 3 −22 8 6 −121 3 2 3
=
1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0
Thus,[v]B = [17, 4,−13]
Since P [v]B = [v]C implies
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 93 / 122
For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.
[B|v] =
1 6 3 2−4 −24 −12 −8
1 5 3 −22 8 6 −121 3 2 3
=
1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0
Thus,[v]B = [17, 4,−13]
Since P [v]B = [v]C implies
[v]C = [2,−2, 3].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 93 / 122
Exercise: For the ordered bases
B =(
2x2 + 3x − 1, 8x2 + x + 1, x2 + 6)
and
C =(
x2 + 3x + 1, 3x2 + 4x + 1, 10x2 + 17x + 5)
of P2, find the transition matrix P from B to C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 94 / 122
Exercise: For the ordered bases
B =(
2x2 + 3x − 1, 8x2 + x + 1, x2 + 6)
and
C =(
x2 + 3x + 1, 3x2 + 4x + 1, 10x2 + 17x + 5)
of P2, find the transition matrix P from B to C.
Answer: P =
20 −30 −6924 −24 −80−9 11 31
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 94 / 122
Exercise: Let P =
2 2 11 −1 21 1 1
be the transition
matrix from B to C. If C =
201
,
120
,
111
, find
the basis B.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 95 / 122
Exercise: Let P =
2 2 11 −1 21 1 1
be the transition
matrix from B to C. If C =
201
,
120
,
111
, find
the basis B.
Answer: B =
633
,
4−13
,
552
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 95 / 122
Exercise: For an ordered basis
B = ([3,−1, 4, 6], [6, 7,−3,−2], [−4,−3, 3, 4], [−2, 0, 1, 2])
of a subspace W of R4, perform the following steps:
1 Use the Simplified Span Method to find asecond ordered basis C.
2 Find the transition matrix P from B to C.3 Find the transition matrix Q from C to B.4 For the given vector v = [10, 14, 3, 12], calculate
[v]B and [v]C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 96 / 122
Section 5.2
The Matrix of a linear transformation:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 97 / 122
Section 5.2
The Matrix of a linear transformation: Let V andW be two finite dimensional real vector spaces suchthat dim(V) = n and dim(W) = m. LetB = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be anordered basis of V and W, respectively.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 97 / 122
Section 5.2
The Matrix of a linear transformation: Let V andW be two finite dimensional real vector spaces suchthat dim(V) = n and dim(W) = m. LetB = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be anordered basis of V and W, respectively.
Let L : V → W be any linear transformation.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 97 / 122
Section 5.2
The Matrix of a linear transformation: Let V andW be two finite dimensional real vector spaces suchthat dim(V) = n and dim(W) = m. LetB = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be anordered basis of V and W, respectively.
Let L : V → W be any linear transformation. Forvj ∈ V, L(vj) ∈ W. For each j, 1 ≤ j ≤ n. Since C isa basis of W, for aij ∈ R, we can write
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 97 / 122
Section 5.2
The Matrix of a linear transformation: Let V andW be two finite dimensional real vector spaces suchthat dim(V) = n and dim(W) = m. LetB = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be anordered basis of V and W, respectively.
Let L : V → W be any linear transformation. Forvj ∈ V, L(vj) ∈ W. For each j, 1 ≤ j ≤ n. Since C isa basis of W, for aij ∈ R, we can write
L(vj) = a1jw1 + a2jw2 + · · · + amjwm
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 97 / 122
Thus, we have
L(v1) = a11w1 + a21w2 + · · +am1wm
L(v2) = a12w1 + a22w2 + · · +am2wm
· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·
L(vn) = a1nw1 + a2nw2 + · · +amnwm
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 98 / 122
Define
ABC =
a11 a12 · · · a1n
a21 a22 · · · a2n... ... ... ...
am1 am2 · · · amn
m×n
.
The matrix ABC is called the matrix of lineartransformation L w.r.t. the bases B and C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 99 / 122
Define
ABC =
a11 a12 · · · a1n
a21 a22 · · · a2n... ... ... ...
am1 am2 · · · amn
m×n
.
The matrix ABC is called the matrix of lineartransformation L w.r.t. the bases B and C.
Remark: ith column of the matrix ABC is [L(vi)]C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 99 / 122
Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 100 / 122
Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT. Then there is a unique m × n matrix ABC suchthat ABC[v]B = [L(v)]C, for all v ∈ V.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 100 / 122
Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT. Then there is a unique m × n matrix ABC suchthat ABC[v]B = [L(v)]C, for all v ∈ V. Furthermore,for 1 ≤ i ≤ n, the ith column of ABC = [L(vi)]C.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 100 / 122
Example: Consider the LT L : P1 → P2, given by
L(p(x)) = xp(x)
with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 101 / 122
Example: Consider the LT L : P1 → P2, given by
L(p(x)) = xp(x)
with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.Compute ABC .
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 101 / 122
Example: Consider the LT L : P1 → P2, given by
L(p(x)) = xp(x)
with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.Compute ABC .
Solution: SinceL(x) = x2 = 1(x2) + 0(x − 1) + 0(x + 1) so that
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 101 / 122
Example: Consider the LT L : P1 → P2, given by
L(p(x)) = xp(x)
with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.Compute ABC .
Solution: SinceL(x) = x2 = 1(x2) + 0(x − 1) + 0(x + 1) so that
[L(x)]C =
100
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 101 / 122
Similarly, L(1) = x = 0(x2) + 1
2(x − 1) + 1
2(x + 1)
implies
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 102 / 122
Similarly, L(1) = x = 0(x2) + 1
2(x − 1) + 1
2(x + 1)
implies
[L(1)]C =
01/21/2
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 102 / 122
Similarly, L(1) = x = 0(x2) + 1
2(x − 1) + 1
2(x + 1)
implies
[L(1)]C =
01/21/2
.
Hence,
ABC =
1 00 1/20 1/2
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 102 / 122
Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R
n andR
m, respectively. Also, let L : Rn → R
m be a LT.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122
Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R
n andR
m, respectively. Also, let L : Rn → R
m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122
Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R
n andR
m, respectively. Also, let L : Rn → R
m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.Form the augmented matrix
[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122
Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R
n andR
m, respectively. Also, let L : Rn → R
m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.Form the augmented matrix
[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]
Apply row reduction on
[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122
Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R
n andR
m, respectively. Also, let L : Rn → R
m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.Form the augmented matrix
[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]
Apply row reduction on
[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)].
to produce [Im | ABC].Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122
Example: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b
with ordered bases B = ([5, 3], [3, 2]) and
C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)
of R2 and P2, respectively.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 104 / 122
Example: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b
with ordered bases B = ([5, 3], [3, 2]) and
C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)
of R2 and P2, respectively. Compute ABC.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 104 / 122
Example: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b
with ordered bases B = ([5, 3], [3, 2]) and
C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)
of R2 and P2, respectively. Compute ABC.
Solution: Since L[5, 3] = 10x2 + 12x + 6
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 104 / 122
Example: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b
with ordered bases B = ([5, 3], [3, 2]) and
C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)
of R2 and P2, respectively. Compute ABC.
Solution: Since L[5, 3] = 10x2 + 12x + 6 andL[3, 2] = 7x2 + 7x + 4. Consider
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 104 / 122
Example: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b
with ordered bases B = ([5, 3], [3, 2]) and
C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)
of R2 and P2, respectively. Compute ABC.
Solution: Since L[5, 3] = 10x2 + 12x + 6 andL[3, 2] = 7x2 + 7x + 4. Consider
3 −2 1 10 7−2 2 −1 12 7
0 −1 1 6 4
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 104 / 122
RREF of the above matrix is
1 0 0 22 140 1 0 62 390 0 1 68 43
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 105 / 122
RREF of the above matrix is
1 0 0 22 140 1 0 62 390 0 1 68 43
so that
ABC =
22 1462 3968 43
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 105 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).
Solution: Standard basis of P3 is
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).
Solution: Standard basis of P3 is {x3, x2, x, 1}.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).
Solution: Standard basis of P3 is {x3, x2, x, 1}. Since
L(x3) = 3x2, L(x2) = 2x, L(x) = 1, L(1) = 0, we have
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).
Solution: Standard basis of P3 is {x3, x2, x, 1}. Since
L(x3) = 3x2, L(x2) = 2x, L(x) = 1, L(1) = 0, we have
ABC =
3 0 0 00 2 0 00 0 1 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 106 / 122
Since
[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B
Since
[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B
Since
[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B
= ABC
4−56−7
=
12−106
Thus, L(4x3 − 5x2 + 6x − 7) = 12x2 − 10x + 6
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 107 / 122
Since
[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B
= ABC
4−56−7
=
12−106
Thus, L(4x3 − 5x2 + 6x − 7) = 12x2 − 10x + 6
Also, note that
L(4x3−5x2+6x−7) = (4x3−5x2+6x−7)′ = 12x2−10x+6.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 107 / 122
Example: Let the matrix of LT L : P1 → P1 with
respect to basis B = (x + 1, x − 1) be[
2 3−1 −2
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 108 / 122
Example: Let the matrix of LT L : P1 → P1 with
respect to basis B = (x + 1, x − 1) be[
2 3−1 −2
]
.
Find the matrix of L with respect to basis C = (x, 1).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 108 / 122
Example: Let the matrix of LT L : P1 → P1 with
respect to basis B = (x + 1, x − 1) be[
2 3−1 −2
]
.
Find the matrix of L with respect to basis C = (x, 1).
Solution: Since ABB =
[
2 3−1 −2
]
, we have
L(x + 1) = 2(x + 1) − 1(x − 1) = x + 3
L(x − 1) = 3(x + 1) − 2(x − 1) = x + 5
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 108 / 122
L(ax + b) = L
(
a + b
2(x + 1) +
a − b
2(x − 1)
)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 109 / 122
L(ax + b) = L
(
a + b
2(x + 1) +
a − b
2(x − 1)
)
L(ax + b) =
(
a + b
2(x + 3) +
a − b
2(x + 5)
)
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 109 / 122
L(ax + b) = L
(
a + b
2(x + 1) +
a − b
2(x − 1)
)
L(ax + b) =
(
a + b
2(x + 3) +
a − b
2(x + 5)
)
so that L(x) = x + 4 and L(1) = −1.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 109 / 122
L(ax + b) = L
(
a + b
2(x + 1) +
a − b
2(x − 1)
)
L(ax + b) =
(
a + b
2(x + 3) +
a − b
2(x + 5)
)
so that L(x) = x + 4 and L(1) = −1.Hence,
ACC =
[
1 04 −1
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 109 / 122
Exercise: Consider the LT L : R3 → R
2, given byL([x, y, z]) = [x + y, y − z].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 110 / 122
Exercise: Consider the LT L : R3 → R
2, given byL([x, y, z]) = [x + y, y − z]. Compute ABC withrespect to bases B = ([1, 0, 1], [0, 1, 1], [1, 1, 1]) andC = ([1, 2], [−1, 1]).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 110 / 122
Exercise: Consider the LT L : R3 → R
2, given byL([x, y, z]) = [x + y, y − z]. Compute ABC withrespect to bases B = ([1, 0, 1], [0, 1, 1], [1, 1, 1]) andC = ([1, 2], [−1, 1]).
Answer: ABC =
[
0 1/3 2/3−1 −2/3 −4/3
]
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 110 / 122
Exercise: Consider the LT L : P3 → M22, given by
L(ax3 + bx2 + cx + d) =
[
−3a − 2c −b + 4d4b − c + 3d −6a − b + 2d
]
.
Compute ABC with respect to standard bases for P3
and M22.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 111 / 122
Exercise: Consider the LT L : P3 → M22, given by
L(ax3 + bx2 + cx + d) =
[
−3a − 2c −b + 4d4b − c + 3d −6a − b + 2d
]
.
Compute ABC with respect to standard bases for P3
and M22.
Answer:
ABC =
−3 0 −2 00 −1 0 40 4 −1 3−6 −1 0 2
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 111 / 122
Exercise: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 112 / 122
Exercise: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.
Compute ABC with respect to bases B = ([5, 3], [3, 2])and C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 112 / 122
Exercise: Consider the LT L : R2 → P2, given by
L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.
Compute ABC with respect to bases B = ([5, 3], [3, 2])and C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1).
Answer:
ABC =
22 1462 3968 43
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 112 / 122
Exercise: Let B = ([1, 2], [2,−1]) andC = ([1, 0], [0, 1]) be ordered bases for R
2. If
L : R2 → R
2 be a LT such that ABC =
[
4 32 −4
]
. Find
L([5, 5]), Also, find L([x, y]) for all [x, y] ∈ R2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 113 / 122
Exercise: Let B = ([1, 2], [2,−1]) andC = ([1, 0], [0, 1]) be ordered bases for R
2. If
L : R2 → R
2 be a LT such that ABC =
[
4 32 −4
]
. Find
L([5, 5]), Also, find L([x, y]) for all [x, y] ∈ R2.
Answer: L([5, 5]) = [15, 2].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 113 / 122
Exercise: Let
B = ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 1, 1], [0, 0, 0, 1]) and
C = ([1, 1, 1], [1, 2, 3], [1, 0, 0])
be ordered bases for R4 and R
3, respectively. If
L : R4 → R
3 be a LT such that ABC =
1 1 0 00 1 1 00 1 0 1
.
Find L?
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 114 / 122
Exercise: Let
B = ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 1, 1], [0, 0, 0, 1]) and
C = ([1, 1, 1], [1, 2, 3], [1, 0, 0])
be ordered bases for R4 and R
3, respectively. If
L : R4 → R
3 be a LT such that ABC =
1 1 0 00 1 1 00 1 0 1
.
Find L?
Answer:
L([x1, x2, x3, x4]) = [−2x1+3x2 +x4, x2 +2x3, x2 +3x3].
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 114 / 122
Matrix for the composition of LinearTransformations:
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 115 / 122
Matrix for the composition of LinearTransformations:
Theorem: Let V1,V2 and V3 be nontrivial finitedimensional vector spaces with ordered bases B, Cand D, respectively.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 115 / 122
Matrix for the composition of LinearTransformations:
Theorem: Let V1,V2 and V3 be nontrivial finitedimensional vector spaces with ordered bases B, Cand D, respectively. Let L1 : V1 → V2 be a lineartransformation with matrix ABC and let L2 : V2 → V3
be a linear transformation with matrix ACD. Thenmatrix
ABD = ACDABC
is the matrix of linear transformationL2 ◦ L1 : V1 → V3 with respect to the bases B and D.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 115 / 122
Example: Let L1 : R2 → R
2 and L2 : R2 → R
3
defined byL1([x, y]) = [y, x]
L2([x, y]) = [x + y, x − y, y]
Find the matrix of L1 and L2 with respect to thestandard basis in each case.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 116 / 122
Example: Let L1 : R2 → R
2 and L2 : R2 → R
3
defined byL1([x, y]) = [y, x]
L2([x, y]) = [x + y, x − y, y]
Find the matrix of L1 and L2 with respect to thestandard basis in each case.Find the matrix of L2 ◦ L1 with respect tostandard basis of R
2 and R3.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 116 / 122
Answer: The matrix of L1 w.r. to B = {[1, 0], [0, 1]} is
ABB =
[
0 11 0
]
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 117 / 122
Answer: The matrix of L1 w.r. to B = {[1, 0], [0, 1]} is
ABB =
[
0 11 0
]
.
The matrix of L2 w. r. to the bases C = [1, 0], [0, 1]and D = {[1, 0, 0], [0, 1, 0], [0, 0, 1]} is
ACD =
1 11 −10 1
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 117 / 122
Thus, the matrix ABD of the linear transformationL2 ◦ L1 : R
2 → R3 w.r. to the bases B and D is
ABD = ACDABB =
1 11 −10 1
[
0 11 0
]
=
1 1−1 11 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 118 / 122
Theorem: Let L : V → W be a linear transformationbetween n-dimensional vector spaces V and W andlet B and C are ordered bases for V and W,respectively. Then L is an isomorphism (orinvertible) if and only if the matrix representationABC for L with respect to B and C is nonsingular.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 119 / 122
Theorem: Let L : V → W be a linear transformationbetween n-dimensional vector spaces V and W andlet B and C are ordered bases for V and W,respectively. Then L is an isomorphism (orinvertible) if and only if the matrix representationABC for L with respect to B and C is nonsingular.
In this case If DCB is the matrix for L−1 with respectto C and B then A−1
BC = DCB.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 119 / 122
Example: Let L1 and L2 be linear operators on R3.
Let
A =
0 −2 10 −1 01 0 0
and B =
1 0 0−2 0 10 −3 0
be matrices for L1 and L2 respectively, with respectto standard basis.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 120 / 122
Example: Let L1 and L2 be linear operators on R3.
Let
A =
0 −2 10 −1 01 0 0
and B =
1 0 0−2 0 10 −3 0
be matrices for L1 and L2 respectively, with respectto standard basis.
Show that L1 and L2 are isomorphisms.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 120 / 122
Example: Let L1 and L2 be linear operators on R3.
Let
A =
0 −2 10 −1 01 0 0
and B =
1 0 0−2 0 10 −3 0
be matrices for L1 and L2 respectively, with respectto standard basis.
Show that L1 and L2 are isomorphisms.
Answer: Since rank(A) = 3 and rank(B) = 3, thematrices A and B are nonsingular.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 120 / 122
Example: Let L1 and L2 be linear operators on R3.
Let
A =
0 −2 10 −1 01 0 0
and B =
1 0 0−2 0 10 −3 0
be matrices for L1 and L2 respectively, with respectto standard basis.
Show that L1 and L2 are isomorphisms.
Answer: Since rank(A) = 3 and rank(B) = 3, thematrices A and B are nonsingular. Hence, L1 and L2
are isomorphisms.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 120 / 122
Find matrices for L−1
1and L−1
2.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 121 / 122
Find matrices for L−1
1and L−1
2.
Answer: Since L−1
1(v) = A−1(v).
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 121 / 122
Find matrices for L−1
1and L−1
2.
Answer: Since L−1
1(v) = A−1(v). Using row
reduction (see Chapter 3), we have
A−1 =
0 0 10 −1 01 −2 0
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 121 / 122
Find matrices for L−1
1and L−1
2.
Answer: Since L−1
1(v) = A−1(v). Using row
reduction (see Chapter 3), we have
A−1 =
0 0 10 −1 01 −2 0
Similarly, L−1
2(v) = B−1(v), where
B−1 =
1 0 00 0 −1/32 1 0
.
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 121 / 122
Thank You
Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 122 / 122