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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 6.3 Extra Topic: Composition and Inverses

Note to Student: In this section we will often use \(U\text{,}\) \(V\) and \(W\) to denote subspaces of \(\R^n\text{,}\) or any other finite-dimensional vector space, such as those we study later on.

Subsection 6.3.1 Composition and Matrix Multiplication

Definition 6.3.1.

Let \(U\text{,}\) \(V\) and \(W\) be vector spaces, and let \(T:U\rightarrow V\) and \(S:V\rightarrow W\) be linear transformations. The composition of \(S\) and \(T\) is the transformation \(S\circ T:U\rightarrow W\) given by
\begin{equation*} (S\circ T)(\mathbf{u})=S(T(\mathbf{u})). \end{equation*}

Example 6.3.2.

Define
\begin{align*} \amp T:\R^2\rightarrow \R^2 \quad\text{by}\quad T\left(\begin{bmatrix}u_1\\u_2\end{bmatrix}\right)=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}, \\ \amp S:\R^2\rightarrow \R^2 \quad\text{by}\quad S\left(\begin{bmatrix}v_1\\v_2\end{bmatrix}\right)=\begin{bmatrix}3v_1-v_2\\-3v_1+v_2\end{bmatrix}. \end{align*}
(You should be able to verify that both transformations are linear.) Examine the effect of \(S\circ T\) on vectors of \(\R^2\text{.}\)
Answer.
From the computational standpoint, the situation is simple.
\begin{align*} (S\circ T)\left(\begin{bmatrix}u_1\\u_2\end{bmatrix}\right)\amp =S\left(T\left(\begin{bmatrix}u_1\\u_2\end{bmatrix}\right)\right)=S\left(\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}\right) \\ \amp =\begin{bmatrix}3(u_1+u_2)-(3u_1+3u_2)\\-3(u_1+u_2)+(3u_1+3u_2)\end{bmatrix} \\ \amp =\mathbf{0}. \end{align*}
This means that \(S\circ T\) maps all vectors of \(\R^2\) to \(\mathbf{0}\text{.}\)
In addition to the computational approach, it is also useful to visualize what happens geometrically. First, observe that
\begin{equation*} T\left(\begin{bmatrix}u_1\\u_2\end{bmatrix}\right)=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}=(u_1+u_2)\begin{bmatrix}1\\3\end{bmatrix}. \end{equation*}
Therefore the image of any vector of \(\R^2\) under \(T\) lies on the line determined by the vector \([1,3]\text{.}\) Even though \(S\) is defined on all of \(\R^2\text{,}\) we are only interested in the action of \(S\) on vectors along the line determined by \([1,3]\text{.}\) Our computations showed that all such vectors map to \(\mathbf{0}\text{.}\) The actions of individual transformations, as well as the composite transformation are shown below.
Composition of functions diagram

Proof.

Let \(T:U\rightarrow V\) and \(S:V\rightarrow W\) be linear transformations. We will show that \(S\circ T\) is linear. For all vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\) of \(U\) and scalars \(a\) and \(b\) we have:
\begin{align*} (S\circ T)(a\mathbf{u}_1+b\mathbf{u}_2)\amp =S(T(a\mathbf{u}_1+b\mathbf{u}_2)) \\ \amp =S(aT(\mathbf{u}_1)+bT(\mathbf{u}_2)) \\ \amp =aS(T(\mathbf{u}_1))+bS(T(\mathbf{u}_2)) \\ \amp =a(S\circ T)(\mathbf{u}_1)+b(S\circ T)(\mathbf{u}_2). \end{align*}

Proof.

For all \(\mathbf{u}\) in \(U\) we have:
\begin{align*} ((R\circ S)\circ T)(\mathbf{u})\amp =(R\circ S)(T(\mathbf{u}))=R(S(T(\mathbf{u}))) \\ \amp =R((S\circ T)(\mathbf{u}))=(R\circ (S\circ T))(\mathbf{u}). \end{align*}
In this section we will consider linear transformations of \(\R^n\) and their standard matrices.

Proof.

For all \(\mathbf{v}\) in \(\R^n\) we have:
\begin{equation*} (S\circ T)(\mathbf{v})=S(T(\mathbf{v}))=S(M_T\mathbf{v})=M_S(M_T\mathbf{v})=(M_SM_T)\mathbf{v}. \end{equation*}

Example 6.3.6.

Example 6.3.2, we discussed a composite transformation \(S\circ T:\R^2\rightarrow \R^2\) given by:
\begin{equation*} T\left(\begin{bmatrix}u_1\\u_2\end{bmatrix}\right)=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}\quad \text{and} \quad S\left(\begin{bmatrix}v_1\\v_2\end{bmatrix}\right)=\begin{bmatrix}3v_1-v_2\\-3v_1+v_2\end{bmatrix}. \end{equation*}
Express \(S\circ T\) as a matrix transformation.
Answer.
The standard matrix for \(T:\R^2\rightarrow \R^2\) is
\begin{equation*} \begin{bmatrix}1\amp 1\\3\amp 3\end{bmatrix} \end{equation*}
and the standard matrix for \(S:\R^2\rightarrow \R^2\) is
\begin{equation*} \begin{bmatrix}3\amp -1\\-3\amp 1\end{bmatrix}. \end{equation*}
The standard matrix for \(S\circ T\) is the product
\begin{equation*} \begin{bmatrix}3\amp -1\\-3\amp 1\end{bmatrix}\begin{bmatrix}1\amp 1\\3\amp 3\end{bmatrix}=\begin{bmatrix}0\amp 0\\0\amp 0\end{bmatrix}. \end{equation*}
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 \(A\text{,}\) \(B\) and \(C\text{,}\) we have \((AB)C=A(BC)\) (see Theorem 4.1.20).
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.

Subsection 6.3.2 Inverses of Linear Transformations

Exploration 6.3.1.

Define a linear transformation \(T:\R^2\rightarrow \R^2\) by \(T(\mathbf{v})=2\mathbf{v}\text{.}\) In other words, \(T\) doubles every vector in \(\R^2\text{.}\) Now define \(S:\R^2\rightarrow \R^2\) by \(S(\mathbf{v})=\frac{1}{2}\mathbf{v}\text{.}\) What happens when we compose these two transformations?
\begin{equation*} (S\circ T)(\mathbf{v})=S(T(\mathbf{v}))=S(2\mathbf{v})=\left(\frac{1}{2}\right)(2)\mathbf{v}=\mathbf{v}, \end{equation*}
\begin{equation*} (T\circ S)(\mathbf{v})=T(S(\mathbf{v}))=T \left (\frac{1}{2}\mathbf{v} \right )=(2)\left(\frac{1}{2}\right)\mathbf{v}=\mathbf{v}. \end{equation*}
Both composite transformations return the original vector \(\mathbf{v}\text{.}\) In other words, \(S\circ T=\id_{\R^2}\) and \(T\circ S=\id_{\R^2}\text{.}\) We say that \(S\) is an inverse of \(T\text{,}\) and \(T\) is an inverse of \(S\text{.}\)

Definition 6.3.8.

Let \(V\) and \(W\) be vector spaces, and let \(T:V\rightarrow W\) be a linear transformation. A transformation \(S:W\rightarrow V\) satisfying \(S\circ T=\id_V\) and \(T\circ S=\id_W\) is called an inverse of \(T\text{.}\) If \(T\) has an inverse, \(T\) is called invertible.

Example 6.3.9.

Let \(T:\R^2\rightarrow \R^2\) be a transformation defined by
\begin{equation*} T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x+y\\x-y\end{bmatrix}\text{.} \end{equation*}
(ow would you verify that \(T\) is linear?). Show that \(S:\R^2\rightarrow \R^2\) given by
\begin{equation*} S\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}0.5x+0.5y\\0.5x-0.5y\end{bmatrix} \end{equation*}
is an inverse of \(T\text{.}\)
Answer.
We will show that \(S\circ T=\id_{\R^2}\text{.}\)
\begin{align*} (S\circ T)\left(\begin{bmatrix}x\\y\end{bmatrix}\right)\amp =S\left(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)\right)=S\left(\begin{bmatrix}x+y\\x-y\end{bmatrix}\right) \\ \amp =\begin{bmatrix}0.5(x+y)+0.5(x-y)\\0.5(x+y)-0.5(x-y)\end{bmatrix} \\ \amp =\begin{bmatrix}x\\y\end{bmatrix}. \end{align*}
We leave it to the reader to verify that \(T\circ S=\id_{\R^2}\text{.}\)
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.

Proof.

Subsection 6.3.3 Linear Transformations of \(\R^n\) and the Standard Matrix of the Inverse Transformation

Every linear transformation \(T:\R^n\rightarrow\R^m\) is a matrix transformation (see Theorem 6.2.24). If \(T\) has an inverse \(S\text{,}\) then by Theorem 6.3.10, \(S\) is also a matrix transformation.
Let \(M_T\) and \(M_S\) denote the standard matrices of \(T\) and \(S\text{,}\) respectively. We see that
\begin{equation*} S\circ T=\id_{\R^n} \text{ and } \circ S=\id_{\R^m} \end{equation*}
if and only if
\begin{equation*} M_SM_T=I_{n} \text{ and } M_TM_S=I_{m}. \end{equation*}
In other words, \(T\) and \(S\) are inverse transformations if and only if \(M_T\) and \(M_S\) are matrix inverses.
Note that if \(S\) is an inverse of \(T\text{,}\) then \(M_T\) and \(M_S\) are square matrices, and \(n=m\text{.}\)

Proof.

Please note that Theorem 6.3.11 is only applicable in the context of linear transformations of \(\R^n\) and their standard matrices. The following example provides us with motivation to investigate inverses further, which we will do in the next section.

Exploration 6.3.2.

Let
\begin{equation*} V=\text{span}\left(\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\1\\1\end{bmatrix}\right). \end{equation*}
Define a linear transformation
\begin{equation*} T:V\rightarrow \R^2 \end{equation*}
by
\begin{equation*} T\left(\begin{bmatrix}1\\0\\0\end{bmatrix}\right)=\begin{bmatrix}1\\1\end{bmatrix}\quad \text{and} \quad T\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right)=\begin{bmatrix}0\\1\end{bmatrix}. \end{equation*}
Observe that
\begin{equation*} \left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\1\\1\end{bmatrix}\right\} \end{equation*}
is a basis of \(V\) (why?). The information about the images of the basis vectors is sufficient to define a linear transformation. This is because every vector \(\mathbf{v}\) in \(V\) can be expressed as a linear combination of the basis elements. The image, \(T(\mathbf{v})\text{,}\) can be found by applying the linearity properties.
At this point we know what transformation \(T\) does, but it is still unclear what the matrix of this linear transformation is. Geometrically speaking, the domain of \(T\) is a plane in \(\R^3\) and its codomain is \(\R^2\text{.}\)
Does \(T\) have an inverse? We are not in a position to answer this question right now because Theorem 6.3.11 does not apply to this situation.

Exercises 6.3.4 Exercises

1.

Let \(T:\R^2\rightarrow \R^2\) and \(S:\R^2\rightarrow \R^2\) be linear transformations with standard matrices
\begin{equation*} M_T=\begin{bmatrix}2\amp -4\\1\amp 2\end{bmatrix}\quad\text{and}\quad M_S=\begin{bmatrix}1\amp -1\\2\amp 1\end{bmatrix} \end{equation*}
respectively. Describe the actions of \(T\text{,}\) \(S\text{,}\) and \(S\circ T\) geometrically, as in Example 6.3.2.

2.

Let \(T:\R^3\rightarrow \R^2\) and \(S:\R^2\rightarrow \R^2\) be linear transformations with standard matrices
\begin{equation*} M_T=\begin{bmatrix}1\amp 0\amp -1\\2\amp 1\amp 0\end{bmatrix}\quad\text{and}\quad M_S=\begin{bmatrix}-1\amp 2\\1\amp -2\end{bmatrix} \end{equation*}
respectively. Describe the actions of \(T\text{,}\) \(S\text{,}\) and \(S\circ T\) geometrically, as in Example 6.3.2.

3.

Complete the Explanation of Example 6.3.9 by verifying that \(T\circ S=\id_{\R^2}\text{.}\)

4.

Let \(T:\R^2\rightarrow \R^2\) be a linear transformation given by
\begin{equation*} T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2x-5y\\-x+3y\end{bmatrix} \end{equation*}
Propose a candidate for the inverse of \(T\) and verify your choice using Definition 6.3.8.
Answer.
\begin{equation*} T^{-1}\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}3x+5y\\x+2y\end{bmatrix} \end{equation*}

5.

Explain why linear transformation \(T:\R^2\rightarrow\R^2\) given by
\begin{equation*} T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2x+2y\\-3x-3y\end{bmatrix}. \end{equation*}
does not have an inverse.

7.

Suppose \(T:U\rightarrow V\) and \(S:V\rightarrow W\) are linear transformations with inverses \(T'\) and \(S'\) respectively. Prove that \(T'\circ S'\) is the inverse of \(S\circ T\text{.}\)
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