Extra Topic: Composition and Inverses

Section 6.3 Extra Topic: Composition and Inverses

Note to Student: In this section we will often use

U,

V

and

W

to denote subspaces of

R^{n},

or any other finite-dimensional vector space, such as those we study later on.

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Subsection 6.3.1 Composition and Matrix Multiplication

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Definition 6.3.1.

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Let

U,

V

and

W

be vector spaces, and let

T : U \to V

and

S : V \to W

be linear transformations. The composition of

S

and

T

is the transformation

S \circ T : U \to W

given by

(S \circ T) (u) = S (T (u)) .

Example 6.3.2.

Define

\begin{aligned} T : R^{2} \to R^{2} by T ([\begin{array}{c} u_{1} \\ u_{2} \end{array}]) = [\begin{array}{c} u_{1} + u_{2} \\ 3 u_{1} + 3 u_{2} \end{array}], \\ S : R^{2} \to R^{2} by S ([\begin{array}{c} v_{1} \\ v_{2} \end{array}]) = [\begin{array}{c} 3 v_{1} - v_{2} \\ - 3 v_{1} + v_{2} \end{array}] . \end{aligned}

(You should be able to verify that both transformations are linear.) Examine the effect of

S \circ T

on vectors of

R^{2} .

Answer.

From the computational standpoint, the situation is simple.

\begin{aligned} (S \circ T) ([\begin{array}{c} u_{1} \\ u_{2} \end{array}]) & = S (T ([\begin{array}{c} u_{1} \\ u_{2} \end{array}])) = S ([\begin{array}{c} u_{1} + u_{2} \\ 3 u_{1} + 3 u_{2} \end{array}]) \\ = [\begin{array}{c} 3 (u_{1} + u_{2}) - (3 u_{1} + 3 u_{2}) \\ - 3 (u_{1} + u_{2}) + (3 u_{1} + 3 u_{2}) \end{array}] \\ = 0 . \end{aligned}

This means that

S \circ T

maps all vectors of

R^{2}

0 .

In addition to the computational approach, it is also useful to visualize what happens geometrically. First, observe that

T ([\begin{matrix} u_{1} \\ u_{2} \end{matrix}]) = [\begin{matrix} u_{1} + u_{2} \\ 3 u_{1} + 3 u_{2} \end{matrix}] = (u_{1} + u_{2}) [\begin{matrix} 1 \\ 3 \end{matrix}] .

Therefore the image of any vector of

R^{2}

under

T

lies on the line determined by the vector

[1, 3] .

Even though

S

is defined on all of

R^{2},

we are only interested in the action of

S

on vectors along the line determined by

[1, 3] .

Our computations showed that all such vectors map to

0 .

The actions of individual transformations, as well as the composite transformation are shown below.

$Composition of functions diagram$

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Theorem 6.3.3.

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The composition of two linear transformations is linear.

Proof.

Let

T : U \to V

and

S : V \to W

be linear transformations. We will show that

S \circ T

is linear. For all vectors

u_{1}

and

u_{2}

U

and scalars

a

and

b

we have:

\begin{aligned} (S \circ T) (a u_{1} + b u_{2}) & = S (T (a u_{1} + b u_{2})) \\ = S (a T (u_{1}) + b T (u_{2})) \\ = a S (T (u_{1})) + b S (T (u_{2})) \\ = a (S \circ T) (u_{1}) + b (S \circ T) (u_{2}) . \end{aligned}

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Theorem 6.3.4.

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Composition of linear transformations is associative. In other words, for linear transformations

T,

S

and

R

$Associativity drawn$

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We have

(R \circ S) \circ T = R \circ (S \circ T) .

Proof.

For all

u

U

we have:

\begin{aligned} ((R \circ S) \circ T) (u) & = (R \circ S) (T (u)) = R (S (T (u))) \\ = R ((S \circ T) (u)) = (R \circ (S \circ T)) (u) . \end{aligned}

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In this section we will consider linear transformations of

R^{n}

and their standard matrices.

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Theorem 6.3.5.

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Let

T : R^{n} \to R^{m}

together with

S : R^{m} \to R^{p}

be linear transformations with standard matrices

M_{T}

and

M_{S},

respectively. Then the composite transformation

S \circ T : R^{n} \to R^{p}

has a standard matrix given by the product

M_{S} M_{T} .

Proof.

For all

v

R^{n}

we have:

(S \circ T) (v) = S (T (v)) = S (M_{T} v) = M_{S} (M_{T} v) = (M_{S} M_{T}) v .

Example 6.3.6.

Example 6.3.2, we discussed a composite transformation

S \circ T : R^{2} \to R^{2}

given by:

T ([\begin{matrix} u_{1} \\ u_{2} \end{matrix}]) = [\begin{matrix} u_{1} + u_{2} \\ 3 u_{1} + 3 u_{2} \end{matrix}] and S ([\begin{matrix} v_{1} \\ v_{2} \end{matrix}]) = [\begin{matrix} 3 v_{1} - v_{2} \\ - 3 v_{1} + v_{2} \end{matrix}] .

Express

S \circ T

as a matrix transformation.

Answer.

The standard matrix for

T : R^{2} \to R^{2}

[\begin{matrix} 1 & 1 \\ 3 & 3 \end{matrix}]

and the standard matrix for

S : R^{2} \to R^{2}

[\begin{matrix} 3 & - 1 \\ - 3 & 1 \end{matrix}] .

The standard matrix for

S \circ T

is the product

[\begin{matrix} 3 & - 1 \\ - 3 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] .

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We conclude this section by revisiting the associative property of matrix multiplication. At the time matrix multiplication was introduced, we skipped the cumbersome proof that for appropriately sized matrices

A,

B

and

C,

we have

(A B) C = A (B C)

(see Theorem 4.1.20).

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We are now in a position to prove this result with ease. Every matrix induces a linear transformation. The product of two matrices can be interpreted as a composition of transformations. Since function composition is associative, so is matrix multiplication. We formalize this observation as a theorem.

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Theorem 6.3.7. Associativity of Matrix Multiplication.

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Let

A,

B

and

C

be matrices of appropriate dimensions so that the product

(A B) C

is defined. Then

(A B) C = A (B C) .

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Subsection 6.3.2 Inverses of Linear Transformations

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Exploration 6.3.1.

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Define a linear transformation

T : R^{2} \to R^{2}

T (v) = 2 v .

In other words,

T

doubles every vector in

R^{2} .

Now define

S : R^{2} \to R^{2}

S (v) = \frac{1}{2} v .

What happens when we compose these two transformations?

(S \circ T) (v) = S (T (v)) = S (2 v) = (\frac{1}{2}) (2) v = v,

(T \circ S) (v) = T (S (v)) = T (\frac{1}{2} v) = (2) (\frac{1}{2}) v = v .

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Both composite transformations return the original vector

v .

In other words,

S \circ T = {id}_{R^{2}}

and

T \circ S = {id}_{R^{2}} .

We say that

S

is an inverse of

T,

and

T

is an inverse of

S .

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Definition 6.3.8.

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Let

V

and

W

be vector spaces, and let

T : V \to W

be a linear transformation. A transformation

S : W \to V

satisfying

S \circ T = {id}_{V}

and

T \circ S = {id}_{W}

is called an inverse of

T .

T

has an inverse,

T

is called invertible.

Example 6.3.9.

Let

T : R^{2} \to R^{2}

be a transformation defined by

T ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} x + y \\ x - y \end{matrix}] .

(ow would you verify that

T

is linear?). Show that

S : R^{2} \to R^{2}

given by

S ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} 0.5 x + 0.5 y \\ 0.5 x - 0.5 y \end{matrix}]

is an inverse of

T .

Answer.

We will show that

S \circ T = {id}_{R^{2}} .

\begin{aligned} (S \circ T) ([\begin{array}{c} x \\ y \end{array}]) & = S (T ([\begin{array}{c} x \\ y \end{array}])) = S ([\begin{array}{c} x + y \\ x - y \end{array}]) \\ = [\begin{array}{c} 0.5 (x + y) + 0.5 (x - y) \\ 0.5 (x + y) - 0.5 (x - y) \end{array}] \\ = [\begin{array}{c} x \\ y \end{array}] . \end{aligned}

We leave it to the reader to verify that

T \circ S = {id}_{R^{2}} .

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Definition 6.3.8 does not specifically require an inverse

S

of a linear transformation

T

to be linear, but it turns out that the requirement that

S \circ T = {id}_{V}

and

T \circ S = {id}_{W}

is sufficient to guarantee that

S

is linear.

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Theorem 6.3.10.

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Suppose

T : V \to W

is an invertible linear transformation. Let

S

be an inverse of

T .

Then

S

is linear.

Proof.

The proof of this result is left to the reader. (See Exercise 6.3.4.6)

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Subsection 6.3.3 Linear Transformations of $R^{n}$ and the Standard Matrix of the Inverse Transformation

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Every linear transformation

T : R^{n} \to R^{m}

is a matrix transformation (see Theorem 6.2.24). If

T

has an inverse

S,

then by Theorem 6.3.10,

S

is also a matrix transformation.

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Let

M_{T}

and

M_{S}

denote the standard matrices of

T

and

S,

respectively. We see that

S \circ T = {id}_{R^{n}} and \circ S = {id}_{R^{m}}

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if and only if

M_{S} M_{T} = I_{n} and M_{T} M_{S} = I_{m} .

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In other words,

T

and

S

are inverse transformations if and only if

M_{T}

and

M_{S}

are matrix inverses.

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Note that if

S

is an inverse of

T,

then

M_{T}

and

M_{S}

are square matrices, and

n = m .

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Theorem 6.3.11.

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Let

T : R^{n} \to R^{n}

be a linear transformation, and let

M

be the standard matrix of

T .

(Existence of Inverses.) $T$ is invertible if and only if $M$ is invertible. If $T$ is invertible, then the inverse is induced by $M^{- 1} .$
(Uniqueness of Inverses.) If $S$ is an inverse of $T,$ then $S$ is unique.