Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations...
Transcript of Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations...
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Linear AlgebraLecture 1 - Solving Linear Equations
Oskar Kędzierski
8 October 2018
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Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
![Page 4: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/4.jpg)
Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
60 problems for classes:http://www.mimuw.edu.pl/~tkozn/ALWNEcw.pdf
![Page 5: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/5.jpg)
Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
60 problems for classes:http://www.mimuw.edu.pl/~tkozn/ALWNEcw.pdf
address: room 5350, the Faculty of Mathematics, Informatics, andMechanics, ul. Banacha 2, Warsaw.
![Page 6: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/6.jpg)
Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
60 problems for classes:http://www.mimuw.edu.pl/~tkozn/ALWNEcw.pdf
address: room 5350, the Faculty of Mathematics, Informatics, andMechanics, ul. Banacha 2, Warsaw.
By R we will denote the real numbers, for example −1, 0,√
2, 3, π ∈ R.
![Page 7: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/7.jpg)
Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
60 problems for classes:http://www.mimuw.edu.pl/~tkozn/ALWNEcw.pdf
address: room 5350, the Faculty of Mathematics, Informatics, andMechanics, ul. Banacha 2, Warsaw.
By R we will denote the real numbers, for example −1, 0,√
2, 3, π ∈ R.By R
n we will denote the n-tuples of real numbers.
![Page 8: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/8.jpg)
Other Details
email: [email protected],website:http://www.mimuw.edu.pl/~oskar/linear_algebra_wne_2018.html
60 problems for classes:http://www.mimuw.edu.pl/~tkozn/ALWNEcw.pdf
address: room 5350, the Faculty of Mathematics, Informatics, andMechanics, ul. Banacha 2, Warsaw.
By R we will denote the real numbers, for example −1, 0,√
2, 3, π ∈ R.By R
n we will denote the n-tuples of real numbers. For example, the3-tuple, (1, −2, 4) ∈ R
3.
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Bibliography
i) K. Hoffman, R. Kunze, Linear Algebra
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Bibliography
i) K. Hoffman, R. Kunze, Linear Algebra
ii) G. Strang, Linear Algebra and its Applications
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Bibliography
i) K. Hoffman, R. Kunze, Linear Algebra
ii) G. Strang, Linear Algebra and its Applications
iii) T. Koźniewski, Wykłady z algebry liniowej (in Polish)
![Page 12: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/12.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
(2, 1){
x1 − x2 = 1x1 + 2x2 = 4
![Page 13: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/13.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
(2, 1){
x1 − x2 = 1x1 + 2x2 = 4
Exactly one solution(2, 1)
![Page 14: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/14.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
{
x1 − x2 = 1x1 − x2 = −3
![Page 15: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/15.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
{
x1 − x2 = 1x1 − x2 = −3
No solutions at all
![Page 16: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/16.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
{
x1 − x2 = 12x1 − 2x2 = 2
![Page 17: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/17.jpg)
You Know Linear Equations Already
1 2 3 4−1−2−3−40
−1
−2
−3
−4
1
2
3
4
x1
x2
{
x1 − x2 = 12x1 − 2x2 = 2
Infinitely manysolutions of the form(x2 + 1, x2), x2 ∈ R
![Page 18: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/18.jpg)
Linear Equations
Linear equation a1x1 + a2x2 + . . . + anxn = b
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Linear Equations
Linear equation a1x1 + a2x2 + . . . + anxn = b in n unknownsx1, . . . , xn
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Linear Equations
Linear equation a1x1 + a2x2 + . . . + anxn = b in n unknownsx1, . . . , xn with the coefficients a1, . . . , an ∈ R
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Linear Equations
Linear equation a1x1 + a2x2 + . . . + anxn = b in n unknownsx1, . . . , xn with the coefficients a1, . . . , an ∈ R and the constant
term b ∈ R.
![Page 22: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/22.jpg)
System of Linear Equations
A system of m linear equations in n unknowns x1, . . . , xn
U :
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2...
.... . .
......
am1x1 + am2x2 + . . . + amnxn = bm
![Page 23: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/23.jpg)
System of Linear Equations
A system of m linear equations in n unknowns x1, . . . , xn
U :
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2...
.... . .
......
am1x1 + am2x2 + . . . + amnxn = bm
with coefficients aij , i = 1, . . . , m, j = 1, . . . , n and constant termsbi ∈ R for i = 1, . . . , m.
![Page 24: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/24.jpg)
System of Linear Equations
A system of m linear equations in n unknowns x1, . . . , xn
U :
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2...
.... . .
......
am1x1 + am2x2 + . . . + amnxn = bm
with coefficients aij , i = 1, . . . , m, j = 1, . . . , n and constant termsbi ∈ R for i = 1, . . . , m. If b1 = b2 = . . . = bm = 0 we call thesystem homogeneous.
![Page 25: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/25.jpg)
Solutions of Systems of Linear Equations
Any n-tuple (x1, . . . , xn) ∈ Rn such that all equations in U are
satisfied is called a solution of the system U.
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Solutions of Systems of Linear Equations
Any n-tuple (x1, . . . , xn) ∈ Rn such that all equations in U are
satisfied is called a solution of the system U. For example, then-tuple (0, . . . , 0) is a solution of a homogenous system of linearequations.
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Solutions of Systems of Linear Equations
Any n-tuple (x1, . . . , xn) ∈ Rn such that all equations in U are
satisfied is called a solution of the system U. For example, then-tuple (0, . . . , 0) is a solution of a homogenous system of linearequations.
A system with no solutions is called inconsistent.
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Solutions of Systems of Linear Equations
Any n-tuple (x1, . . . , xn) ∈ Rn such that all equations in U are
satisfied is called a solution of the system U. For example, then-tuple (0, . . . , 0) is a solution of a homogenous system of linearequations.
A system with no solutions is called inconsistent. Two systems oflinear equations are called equivalent if they have the same sets ofsolutions.
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Operations on Equations
Any equation a1x1 + a2x2 + . . . + anxn = b can be multiplied by anon-zero constant c ∈ R − {0} in order to get the equationca1x1 + ca2x2 + . . . + canxn = cb.
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Operations on Equations
Any equation a1x1 + a2x2 + . . . + anxn = b can be multiplied by anon-zero constant c ∈ R − {0} in order to get the equationca1x1 + ca2x2 + . . . + canxn = cb.
One can add any two equationsa1x1 + a2x2 + . . . + anxn = b, a′
1x1 + a′
2x2 + . . . + a′
nxn = b′ and getthe equation (a1 + a′
1)x1 + (a2 + a′
2)x2 + . . . + (an + a′
n)xn = b + b′.
![Page 31: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/31.jpg)
Equivalent System of Linear Equations
TheoremThe following operations on a system of linear equations do notchange the set of its solutions (i.e. they lead to an equivalentsystem):
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Equivalent System of Linear Equations
TheoremThe following operations on a system of linear equations do notchange the set of its solutions (i.e. they lead to an equivalentsystem):
i) swapping the order of any two equations,
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Equivalent System of Linear Equations
TheoremThe following operations on a system of linear equations do notchange the set of its solutions (i.e. they lead to an equivalentsystem):
i) swapping the order of any two equations,
ii) multiplying any equation by a non-zero constant,
![Page 34: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/34.jpg)
Equivalent System of Linear Equations
TheoremThe following operations on a system of linear equations do notchange the set of its solutions (i.e. they lead to an equivalentsystem):
i) swapping the order of any two equations,
ii) multiplying any equation by a non-zero constant,
iii) adding an equation to the other.
![Page 35: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/35.jpg)
Equivalent System of Linear Equations
TheoremThe following operations on a system of linear equations do notchange the set of its solutions (i.e. they lead to an equivalentsystem):
i) swapping the order of any two equations,
ii) multiplying any equation by a non-zero constant,
iii) adding an equation to the other.
Proof.Any solution of the original system is a solution of the new system.All above operations are reversible.
![Page 36: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/36.jpg)
A General Solution
A general solution of the system of linear equation U is anequivalent linear system U ′ of the form:
U ′ :
xj1 = c11x1 + c12x2 + . . . + c1nxn +d1
xj2 = c21x1 + c22x2 + . . . + c2nxn +d2...
.... . .
......
xjk = ck1x1 + ck2x2 + . . . + cknxn +dk
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A General Solution
A general solution of the system of linear equation U is anequivalent linear system U ′ of the form:
U ′ :
xj1 = c11x1 + c12x2 + . . . + c1nxn +d1
xj2 = c21x1 + c22x2 + . . . + c2nxn +d2...
.... . .
......
xjk = ck1x1 + ck2x2 + . . . + cknxn +dk
where {j1, . . . , jk} ⊂ {1, . . . , n}, j1 < j2 < . . . < jk and cij = 0 forany i = 1, . . . , k and j = j1, . . . , jk . That is, the unknownsxj1 , . . . , xkk
appear only on the left hand-side of each equationexactly once.
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A General Solution
A general solution of the system of linear equation U is anequivalent linear system U ′ of the form:
U ′ :
xj1 = c11x1 + c12x2 + . . . + c1nxn +d1
xj2 = c21x1 + c22x2 + . . . + c2nxn +d2...
.... . .
......
xjk = ck1x1 + ck2x2 + . . . + cknxn +dk
where {j1, . . . , jk} ⊂ {1, . . . , n}, j1 < j2 < . . . < jk and cij = 0 forany i = 1, . . . , k and j = j1, . . . , jk . That is, the unknownsxj1 , . . . , xkk
appear only on the left hand-side of each equationexactly once.
The unknowns xj1 , . . . , xjk are called basic (or dependent)variables. The other unknowns are called free variables orparameters.
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Matrices
A m × n matrix D with entries in R is a rectangular array of realnumbers arranged in m rows and n columns, i.e.
D =
d11 d12 . . . d1n
d21 d22 . . . d2n
......
. . ....
dm1 dm2 . . . dmn
where dij ∈ R.
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Matrices
A m × n matrix D with entries in R is a rectangular array of realnumbers arranged in m rows and n columns, i.e.
D =
d11 d12 . . . d1n
d21 d22 . . . d2n
......
. . ....
dm1 dm2 . . . dmn
where dij ∈ R. Sometimes we write D = [dij ] for i = 1, . . . , m,j = 1, . . . , n.
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Matrices
A m × n matrix D with entries in R is a rectangular array of realnumbers arranged in m rows and n columns, i.e.
D =
d11 d12 . . . d1n
d21 d22 . . . d2n
......
. . ....
dm1 dm2 . . . dmn
where dij ∈ R. Sometimes we write D = [dij ] for i = 1, . . . , m,j = 1, . . . , n. The set of all m-by-n matrices with entries in R willbe denoted M(m × n;R).
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Matrix of a System of Linear Equations
To every system of linear equations
U :
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2...
.... . .
......
am1x1 + am2x2 + . . . + amnxn = bm
we associate its m × (n + 1) matrix
a11 a12 . . . a1n b1
a21 a22 . . . a2n b2...
.... . .
......
am1 am2 . . . amn bm
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Matrix of a System of Linear Equations
The submatrix
a11 a12 . . . a1n
a21 a22 . . . a2n
......
. . ....
am1 am2 . . . amn
is called the matrix of coefficients. The last column
b1
b2...
bm
consists of constant terms.
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Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
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Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
i) swapping any two rows of the matrix,
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Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
i) swapping any two rows of the matrix,
ii) multiplying any row by a non-zero constant c , i.e. replacingthe i-th row [ai1 ai2 . . . ain] with the row [cai1 cai2 . . . cain],
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Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
i) swapping any two rows of the matrix,
ii) multiplying any row by a non-zero constant c , i.e. replacingthe i-th row [ai1 ai2 . . . ain] with the row [cai1 cai2 . . . cain],
iii) adding any row to the other, i.e. replacing the i-th row[ai1 ai2 . . . ain] with the row [ai1 + aj1 ai2 + aj2 . . . ain + ajn].
![Page 48: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/48.jpg)
Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
i) swapping any two rows of the matrix,
ii) multiplying any row by a non-zero constant c , i.e. replacingthe i-th row [ai1 ai2 . . . ain] with the row [cai1 cai2 . . . cain],
iii) adding any row to the other, i.e. replacing the i-th row[ai1 ai2 . . . ain] with the row [ai1 + aj1 ai2 + aj2 . . . ain + ajn].
By the Theorem the elementary row operations lead to a matrix ofan equivalent linear system.
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Elementary Row Operations
There are three elementary row operations on a matrix of a linearsystem:
i) swapping any two rows of the matrix,
ii) multiplying any row by a non-zero constant c , i.e. replacingthe i-th row [ai1 ai2 . . . ain] with the row [cai1 cai2 . . . cain],
iii) adding any row to the other, i.e. replacing the i-th row[ai1 ai2 . . . ain] with the row [ai1 + aj1 ai2 + aj2 . . . ain + ajn].
By the Theorem the elementary row operations lead to a matrix ofan equivalent linear system. The algorithm using the threeelementary row operations, leading to a general solution is calledthe Gaussian elimination.
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The (Reduced) Echelon Form
The leading coefficient (or pivot) of a non-zero row is theleftmost non-zero entry of the row.
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The (Reduced) Echelon Form
The leading coefficient (or pivot) of a non-zero row is theleftmost non-zero entry of the row.
A matrix is in an echelon form if:
i) all non-zero rows are above all zero rows,
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The (Reduced) Echelon Form
The leading coefficient (or pivot) of a non-zero row is theleftmost non-zero entry of the row.
A matrix is in an echelon form if:
i) all non-zero rows are above all zero rows,
ii) the leading coefficient of any row lies strictly to the right ofthe leading coefficient of any upper row.
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The (Reduced) Echelon Form
The leading coefficient (or pivot) of a non-zero row is theleftmost non-zero entry of the row.
A matrix is in an echelon form if:
i) all non-zero rows are above all zero rows,
ii) the leading coefficient of any row lies strictly to the right ofthe leading coefficient of any upper row.
A matrix is in a reduced echelon form if it is in an echelon form,all leading coefficients are equal to 1 and every leading coefficientis the only non-zero element in its column.
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Example
The following matrix is in an echelon form. The leadingcoefficients are marked with circles.
0 1© 2 0 3 2 50 0 1© 2 1 0 00 0 0 0 1© 2 60 0 0 0 0 0 00 0 0 0 0 0 0
It is not in a reduced echelon form because in columns 3 and 5there are leading coefficients and other non-zero entries.
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The Gaussian Elimination
TheoremAny matrix can be brought into a reduced echelon form usingelementary row operations.
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The Gaussian Elimination
TheoremAny matrix can be brought into a reduced echelon form usingelementary row operations.
Proof.Use induction on the number of columns to prove that every matrixcan be brought into an echelon form.
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The Gaussian Elimination
TheoremAny matrix can be brought into a reduced echelon form usingelementary row operations.
Proof.Use induction on the number of columns to prove that every matrixcan be brought into an echelon form. Let A = [aij ] ∈ M(m × 1;R).
If A 6=
0...0
and, for example, a11 6= 0 then
a11...
am1
r2−a21a11
r1
...rm−
am1a11
r1
−→
a11
0...0
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The Gaussian Elimination
Proof.Let A = [aij ] ∈ M(m × n;R) and let n > 1.
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The Gaussian Elimination
Proof.Let A = [aij ] ∈ M(m × n;R) and let n > 1. Let k ∈ N be thenumber of first non-zero column, changing the order of rows onecan assume that a1k 6= 0.
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The Gaussian Elimination
Proof.Let A = [aij ] ∈ M(m × n;R) and let n > 1. Let k ∈ N be thenumber of first non-zero column, changing the order of rows onecan assume that a1k 6= 0. Then
0 · · · 0 a1k a1(k+1) · · · a1n
0 · · · 0 a2k a2(k+1) · · · a2n
.... . .
......
.... . .
...0 · · · 0 amk am(k+1) · · · amn
r2−a2ka1k
r1
...rm−
amka1k
r1
−→
0 · · · 0 a1k a1(k+1) · · · a1n
0 · · · 0 0 b2(k+1) · · · b2n
.... . .
......
.... . .
...0 · · · 0 0 bm(k+1) · · · bmn
for some bij ∈ R.
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The Gaussian Elimination
Proof.By the inductive assumption the matrix in the lower right corner,i.e.
b2(k+1) · · · b2n
.... . .
...bm(k+1) · · · bmn
can be brought to an echelon form by elementary operations.
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The Gaussian Elimination
Proof.By the inductive assumption the matrix in the lower right corner,i.e.
b2(k+1) · · · b2n
.... . .
...bm(k+1) · · · bmn
can be brought to an echelon form by elementary operations. Thesame operations will bring matrix
0 · · · 0 a1k a1(k+1) · · · a1n
0 · · · 0 0 b2(k+1) · · · b2n
0. . . 0
......
. . ....
0 · · · 0 0 bm(k+1) · · · bmn
to an echelon form.
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The Gaussian Elimination
Proof.Assume that matrix A = [aij ] ∈ M(m × n;R) is in echelon formand the leading coefficients are a1j1 , a2j2 , . . . , am′jm′
wherej1 < j2 < . . . < jm′ and m′ ≤ m, i.e. rows m′ + 1, m′ + 2, . . . , m arezero.
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The Gaussian Elimination
Proof.Assume that matrix A = [aij ] ∈ M(m × n;R) is in echelon formand the leading coefficients are a1j1 , a2j2 , . . . , am′jm′
wherej1 < j2 < . . . < jm′ and m′ ≤ m, i.e. rows m′ + 1, m′ + 2, . . . , m arezero.
0 a1j1 ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 a2j2 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 0 0 a3j3 · · · ∗ ∗ ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 am′jm′∗ . . . ∗
0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
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The Gaussian Elimination
Proof.The following elementary operations will bring the matrix A in anechelon form into a reduced echelon form
rk − akji
aiji
ri for i = 2, . . . , m′, k = 1, . . . , i − 1,
ri/aiji for i = 1, . . . , m′.
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The Gaussian Elimination
Proof.The following elementary operations will bring the matrix A in anechelon form into a reduced echelon form
rk − akji
aiji
ri for i = 2, . . . , m′, k = 1, . . . , i − 1,
ri/aiji for i = 1, . . . , m′.
In short, in each of the column j1, j2, . . . , jm′ we use the leadingcoefficient to make the entries above it zero and then we divide thecorresponding row to make the leading coefficient equal to 1.
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The Gaussian Elimination
Proof.The following elementary operations will bring the matrix A in anechelon form into a reduced echelon form
rk − akji
aiji
ri for i = 2, . . . , m′, k = 1, . . . , i − 1,
ri/aiji for i = 1, . . . , m′.
In short, in each of the column j1, j2, . . . , jm′ we use the leadingcoefficient to make the entries above it zero and then we divide thecorresponding row to make the leading coefficient equal to 1.
0 a1j1 ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 a2j2 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 0 0 a3j3 · · · ∗ ∗ ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 am′jm′∗ . . . ∗
0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
r1/a1j1−→
0 1 ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 a2j2 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 0 0 a3j3 · · · ∗ ∗ ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 am′jm′∗ . . . ∗
0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
r1−
a1j2a2j2
r2
−→
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The Gaussian Elimination
Proof.
0 1 0 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 a2j2 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 0 0 a3j3 · · · ∗ ∗ ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 am′jm′∗ . . . ∗
0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
r2/a2j2−→
0 1 0 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 1 ∗ ∗ · · · ∗ ∗ ∗ ∗ ∗0 0 0 0 a3j3 · · · ∗ ∗ ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 am′jm′∗ . . . ∗
0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
r1−
a1j3a3j3
r3
r2−
a2j3a3j3
r3
−→ · · ·
. . . −→
0 1 0 ∗ 0 · · · ∗ 0 ∗ ∗ ∗0 0 1 ∗ 0 · · · ∗ 0 ∗ ∗ ∗0 0 0 0 1 · · · ∗ 0 ∗ . . . ∗...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 1 ∗ . . . ∗0 0 0 0 0 · · · 0 0 0 0 0...
......
......
. . ....
......
. . ....
0 0 0 0 0 · · · 0 0 0 0 0
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The Gaussian Elimination
How to solve a system of linear equations?
Bring a matrix of a system of linear equation to a reduced echelonform. If there is a pivot in the column of constant terms thesystem if inconsistent. Otherwise, the general solution can be readfrom the echelon form by choosing the basic variables as thosecorresponding to columns with a pivot.
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Example
Let’s solve the system
{
x1 − 2x2 + x3 − x4 = 22x1 − 4x2 + 3x3 + x4 = 0
The matrix of this system is
[
1 −2 1 −1 22 −4 3 1 0
]
By the elementary row operation r2 − 2r1 we put the matrix in anechelon form, i.e.
[
1 −2 1 −1 20 0 1 3 −4
]
The elementary operation r1 − r2 puts matrix in a reduced echelonform, that is
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Example (continued)
[
1© −2 0 −4 60 0 1© 3 −4
]
There is no leading coefficient in the constant term columnn so ithas solutions. The basic variables are x1, x3 and the free variablesare x2, x4.
The general solution is
{
x1 = 2x2 + 4x4 + 6x3 = − 3x4 − 4
, x2, x4 ∈ R.
Every solution of this linear system is of the form
(2x2 + 4x4 + 6, x2, −3x4 − 4, x4), x2, x4 ∈ R.
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The Uniqueness of the Reduced Echelon Form
Proposition
Let A ∈ M(m × n;R) be a matrix. If matrices B, C ∈ M(m × n;R)were obtained from A by a series of elementary row operations andthey are in a reduced echelon form then B = C.
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The Uniqueness of the Reduced Echelon Form
Proposition
Let A ∈ M(m × n;R) be a matrix. If matrices B, C ∈ M(m × n;R)were obtained from A by a series of elementary row operations andthey are in a reduced echelon form then B = C.
Proof.Let j be the number of the leftmost column where the matrices Band C differ. Let
1 ≤ j1 < j2 < . . . < jk < j ,
be the numbers of the columns with pivots in B and C smallerthan j . Let B′ and C ′ be submatrices of matrices B and C ,respectively, consisting of columns j1, . . . , jk , j . Let UB , UC besystems of linear equations which matrices are equal to B′ and C ′,respectively (the last column consists of constant terms). By theassumption, the systems UB and UC are equivalent.
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The Uniqueness of the Reduced Echelon Form (continued)
Proof.
B′ =
1 0 · · · 0 b1j
0 1... b2j
.... . . 0
...0 0 · · · 1 bkj
0 0 · · · 0 b(k+1)j...
. . . 0...
0 0 · · · 0 bmj
, C ′ =
1 0 · · · 0 c1j
0 1... c2j
.... . . 0
...0 0 · · · 1 ckj
0 0 · · · 0 c(k+1)j...
. . . 0...
0 0 · · · 0 cmj
If b(k+1)j = 0 then bij = 0 for i ≥ k + 1 and if c(k+1)j = 0 thencij = 0 for i ≥ k + 1.
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The Uniqueness of the Reduced Echelon Form (continued)
Proof.
B′ =
1 0 · · · 0 b1j
0 1... b2j
.... . . 0
...0 0 · · · 1 bkj
0 0 · · · 0 b(k+1)j...
. . . 0...
0 0 · · · 0 bmj
, C ′ =
1 0 · · · 0 c1j
0 1... c2j
.... . . 0
...0 0 · · · 1 ckj
0 0 · · · 0 c(k+1)j...
. . . 0...
0 0 · · · 0 cmj
If b(k+1)j = 0 then bij = 0 for i ≥ k + 1 and if c(k+1)j = 0 thencij = 0 for i ≥ k + 1. If b(k+1)j = 0, c(k+1)j 6= 0 orb(k+1)j 6= 0, c(k+1)j = 0 then one of the systems UB , UC isinconsistent and the other is consistent, which leads to acontradiction.
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The Uniqueness of the Reduced Echelon Form (continued)
Proof.
B′ =
1 0 · · · 0 b1j
0 1... b2j
.... . . 0
...0 0 · · · 1 bkj
0 0 · · · 0 b(k+1)j...
. . . 0...
0 0 · · · 0 bmj
, C ′ =
1 0 · · · 0 c1j
0 1... c2j
.... . . 0
...0 0 · · · 1 ckj
0 0 · · · 0 c(k+1)j...
. . . 0...
0 0 · · · 0 cmj
If b(k+1)j = 0 then bij = 0 for i ≥ k + 1 and if c(k+1)j = 0 thencij = 0 for i ≥ k + 1. If b(k+1)j = 0, c(k+1)j 6= 0 orb(k+1)j 6= 0, c(k+1)j = 0 then one of the systems UB , UC isinconsistent and the other is consistent, which leads to acontradiction. The j−th column cannot contain a pivot in B andC (then the pivot would be in (k + 1)-th row and the j-th columnwould be the same for B and C), therefore b(k+1)j 6= 0, c(k+1)j 6= 0is not possible. The remaining case is b(k+1)j = 0, c(k+1)j = 0 inwhich the system UB has a unique solution (b1j , b2j , . . . , bkj) andthe system UC has a unique solution (c1j , c2j , . . . , ckj ), which againleads to a contradiction.
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The Uniqueness of the Reduced Echelon Form (continued)
RemarkObviously, echelon form is not unique.
[
1 1 11 2 1
]
r2−r1−→[
1 1 10 1 0
]
[
1 1 11 2 1
]
r1↔r2−→[
1 2 11 1 1
]
r2−r1−→[
1 2 10 −1 0
]
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The Uniqueness of the Reduced Echelon Form (continued)
RemarkObviously, echelon form is not unique.
[
1 1 11 2 1
]
r2−r1−→[
1 1 10 1 0
]
[
1 1 11 2 1
]
r1↔r2−→[
1 2 11 1 1
]
r2−r1−→[
1 2 10 −1 0
]
The reduced echelon form of matrix
[
1 1 11 2 1
]
is
[
1 0 10 1 0
]
.
![Page 79: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/79.jpg)
The Uniqueness of the Reduced Echelon Form (continued)
RemarkOne can read a general solution from a matrix which after apermutation (i.e. change of the order) of columns is in the reducedechelon form (by choosing basic variables as those correspondingto columns which after the permutation contain a pivot).
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The Uniqueness of the Reduced Echelon Form (continued)
RemarkOne can read a general solution from a matrix which after apermutation (i.e. change of the order) of columns is in the reducedechelon form (by choosing basic variables as those correspondingto columns which after the permutation contain a pivot).
A general solution of the system
[
2 1© 3 0 2−1 0 5 1© −7
]
is{
x2 = −2x1 − 3x3 + 2x4 = x1 − 5x3 − 7
, x1, x3 ∈ R.
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Reduced Echelon Form of a Square Matrix
Proposition
Let A ∈ M(n × n;R) be a square matrix (i.e. it has the n rows andn columns). Then the reduced echelon form of A either has a zero
row or it is equal to In =
1 0. . .
0 1
∈ M(n × n;R).
![Page 82: Linear Algebra - mimuwoskar/lecture_1.pdf · Linear Algebra Lecture 1 - Solving Linear Equations Oskar Kędzierski 8 October 2018](https://reader031.fdocuments.in/reader031/viewer/2022021723/5c757eaf09d3f2d3778b809d/html5/thumbnails/82.jpg)
Reduced Echelon Form of a Square Matrix
Proposition
Let A ∈ M(n × n;R) be a square matrix (i.e. it has the n rows andn columns). Then the reduced echelon form of A either has a zero
row or it is equal to In =
1 0. . .
0 1
∈ M(n × n;R).
Proof.By definition, the numbers of columns with pivots form a strictlyincreasing sequence
1 ≤ j1 < j2 < . . . < jk ≤ n.
Therefore k ≤ n. If k < n then there are n − k zero rows (only krows contain a pivot).
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Reduced Echelon Form of a Square Matrix
Proposition
Let A ∈ M(n × n;R) be a square matrix (i.e. it has the n rows andn columns). Then the reduced echelon form of A either has a zero
row or it is equal to In =
1 0. . .
0 1
∈ M(n × n;R).
Proof.By definition, the numbers of columns with pivots form a strictlyincreasing sequence
1 ≤ j1 < j2 < . . . < jk ≤ n.
Therefore k ≤ n. If k < n then there are n − k zero rows (only krows contain a pivot). The case k = n is possible only ifj1 = 1, j2 = 2, . . . , jn = n, i.e. the reduced form of A is equal toIn.