2.8 Composition and Invertibility of Linear Transformations The standard matrix of a linear transformation T can be used to find a generating set for the range of T.

1

Proof

Property The range of a linear transformation equals the span of the columns of its

standard matrix.

2

Range

images

fS1 S2

Definition

Range

fS1 S2

not onto

onto

3

Definition

(Review) Theorem 1.6

Proof

Theorem 2.10

4

Example: Is T onto?

⇒ with the reduced row echelon form

The system of linear equations Ax = b may be written as TA(x) = b. Ax = b has a solution if and only if b is in the range of TA. TA is not onto if and only if ∃ b such that Ax = b is inconsistent.

T : R3 ! R3 with T

0

@

2

4x1

x2

x3

3

5

1

A =

2

4x1 + 2x2 + 4x3

x1 + 3x2 + 6x3

2x1 + 5x2 + 10x3

3

5

5

Definition

(Review) Theorem 1.6

Proof

6

one-to-one not one-to-one

An equivalent condition for a function f to be one-to-one is that f (u) = f (v) implies u = v.

0 • • 0 0 • • 0

z •

Definition

7

one-to-one not one-to-one

0 • • 0 0 • • 0

z •

Definition

Definition

Property

8

Example:

Find a generating set for the null space of T.

is the standard matrix of T, and has the reduced row echelon form

The reduced row echelon form corresponds to the linear equations

so a generating set for the null space of T is { [ 1 1 0 ]T }

the null space of T is the set of solutions to Ax = 0, where

T : R3 ! R2 with T

0

@

2

4x1

x2

x3

3

5

1

A =

x1 � x2 + 2x3

�x1 + x2 � 3x3

�

x1 � x2 = 0x3 = 0

9

Proof

Theorem 2.11

(Review) Theorem 1.8

10

Example:

The standard matrix A has rank 2 (nullity 1), so T is not one-to-one.

If TA is one-to-one and the solution of Ax = b exists, then the solution is unique. Conversely, if there is at most one solution to Ax = b for every b, then TA is one-to-one.

Example: row reduced echelon form

rank A = 3 and nullity A = 2. ⇒ TA is not onto and not one-to-one. ⇒ Ax = b is inconsistent for b not in the range of TA, and the solutions of Ax = b is never unique.

T : R3 ! R3 with T

0

@

2

4x1

x2

x3

3

5

1

A =

2

4x1 + 2x2 + 4x3

x1 + 3x2 + 6x3

2x1 + 5x2 + 10x3

3

5 = Ax

A =

2

664

0 0 1 3 32 3 1 5 24 6 1 6 24 6 1 7 1

3

775R =

2

664

1 1.5 0 1 00 0 1 3 00 0 0 0 10 0 0 0 0

3

775

11

Definition.

Example:

⇒

f

✓x1

x2

�◆=

2

4x

21

x1x2

x1 + x2

3

5g

0

@

2

4x1

x2

x3

3

5

1

A =

x1 � x3

3x2

�

(g � f)✓

x1

x2

�◆=

x

21 � (x1 + x2)

3x1x2

�

f gS1 S2 S3

uf(u) g(f(u))· · ·g � f

12

Example:

⇒ TB : reflection about the x-axis in R2.

⇒ TA : rotation by 180 in R2.

⇒ TBA : reflection about the y-axis in R2 = rotation by 180 followed by reflection about the x-axis in R2.

Proof

A

A B

Theorem 2.12

A =

�1 00 �1

�

B =

1 00 �1

�

TBA

✓u1

u2

�◆= (BA)

u1

u2

�=

�1 00 1

� u1

u2

�=

�u1

u2

�

13

Definition.

If A ∈ Rn×n is invertible, then for all v ∈ Rn we have TA(TA-1(v)) = (TATA-1) (v) = TAA-1(v) = TIn

(v) = Inv = v, and TA-1(TA(v)) = v. Thus TA

-1 = TA-1 .

Example:

A =

1 23 5

�A�1 =

�5 23 �1

�

T�1A

✓v1v2

�◆= TA�1

✓v1v2

�◆=

�5 23 �1

� v1v2

�=

�5v1 + 2v23v1 � v2

�

TA

✓v1v2

�◆=

v1 + 2v23v1 + 5v2

�

14

Proof

Theorem 2.13

15

Property of T The number of solutions of Ax=b

Property of the columns of A

Property of the rank of A

T is onto Ax = b has at least one solution for every b in Rm.

The columns of A are a generating set for Rm.

rank A = m

T is one-to-one Ax = b has at most one solution for every b in Rm.

The columns of A are linearly independent.

rank A = n

T is invertible Ax = b has a unique solution for every b in Rm

The columns of A are a linearly independent generating set for Rm.

rank A = m = n

Theorem

2.10

2.11

2.13

16

Section 2.8: Problems 1, 5, 13, 17, 19, 25, 29, 33, 37

Homework Set for Section 2.8

17