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