Note to Student: In this section we will often use \(U\), \(V\) and \(W\) to denote subspaces of \(\RR ^n\), or any other finite-dimensional vector space, such as those we study in Vector Spaces.

Composition and Inverses of Linear Transformations

Composition of Linear Transformations

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 \(\vec {u}_1\) and \(\vec {u}_2\) of \(U\) and scalars \(a\) and \(b\) we have:

\begin{align*} (S\circ T)(a\vec {u}_1+b\vec {u}_2)&=S(T(a\vec {u}_1+b\vec {u}_2))\\ &=S(aT(\vec {u}_1)+bT(\vec {u}_2))\\ &=aS(T(\vec {u}_1))+bS(T(\vec {u}_2))\\ &=a(S\circ T)(\vec {u}_1)+b(S\circ T)(\vec {u}_2) \end{align*}

For all \(\vec {u}\) in \(U\) we have:

\begin{align*} ((R\circ S)\circ T)(\vec {u})&=(R\circ S)(T(\vec {u}))=R(S(T(\vec {u})))\\ &=R((S\circ T)(\vec {u}))=(R\circ (S\circ T))(\vec {u}) \end{align*}

Composition and Matrix Multiplication

In this section we will consider linear transformations of \(\RR ^n\) and their standard matrices.

For all \(\vec {v}\) in \(\RR ^n\) we have:

\[(S\circ T)(\vec {v})=S(T(\vec {v}))=S(M_T\vec {v})=M_S(M_T\vec {v})=(M_SM_T)\vec {v}\]

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 \((AB)C=A(BC)\). (See Theorem ??.) 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.

Inverses of Linear Transformations

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

Linearity of Inverses of Linear Transformations

Definition 13 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.

The proof of this result is left to the reader. (See Practice Problem ??)

Linear Transformations of \(\RR ^n\) and the Standard Matrix of the Inverse Transformation

Every linear transformation \(T:\RR ^n\rightarrow \RR ^m\) is a matrix transformation. (See Theorem ??.) If \(T\) has an inverse \(S\), then by Theorem 16, \(S\) is also a matrix transformation. Let \(M_T\) and \(M_S\) denote the standard matrices of \(T\) and \(S\), respectively. We see that \(S\circ T=\id _{\RR ^n}\) and \(T\circ S=\id _{\RR ^m}\) if and only if \(M_SM_T=I_{n}\) and \(M_TM_S=I_{m}\). 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 \(T\), then \(M_T\) and \(M_S\) are square matrices, and \(n=m\).

Part 1 follows directly from the preceding discussion. Part 2 follows from uniqueness of matrix inverses. (Theorem ??.)

Please note that Theorem 18 is only applicable in the context of linear transformations of \(\RR ^n\) and their standard matrices. The following example provides us with motivation to investigate inverses further, which we will do in Existence of the Inverse of a Linear Transformation.

Let
\[V=\text {span}\left (\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\1\\1\end{bmatrix}\right )\]
Define a linear transformation
\[T:V\rightarrow \RR ^2\]
by
\[T\left (\begin{bmatrix}1\\0\\0\end{bmatrix}\right )=\begin{bmatrix}1\\1\end{bmatrix}\quad \text {and} \quad T\left (\begin{bmatrix}1\\1\\1\end{bmatrix}\right )=\begin{bmatrix}0\\1\end{bmatrix}\]
Observe that \(\left \{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\1\\1\end{bmatrix}\right \}\) 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 \(\vec {v}\) in \(V\) can be expressed as a linear combination of the basis elements in a unique way. The image, \(T(\vec {v})\), 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 \(\RR ^3\) and its codomain is \(\RR ^2\).

Does \(T\) have an inverse? We are not in a position to answer this question right now because Theorem 18 does not apply to this situation.

Exploration highlights the necessity of further discussion of inverses and of matrices associated with linear transformations. In Example ??, we will prove that \(T\) has an inverse, and in Example ??, we will address the issue of matrices associated with \(T\) and its inverse.

Practice Problems

Let \(T:\RR ^2\rightarrow \RR ^2\) and \(S:\RR ^2\rightarrow \RR ^2\) be linear transformations with standard matrices
\[M_T=\begin{bmatrix}2&-4\\1&2\end{bmatrix}\quad \text {and}\quad M_S=\begin{bmatrix}1&-1\\2&1\end{bmatrix}\]
respectively. Describe the actions of \(T\), \(S\), and \(S\circ T\) geometrically, as in Example 2.
Let \(T:\RR ^3\rightarrow \RR ^2\) and \(S:\RR ^2\rightarrow \RR ^2\) be linear transformations with standard matrices
\[M_T=\begin{bmatrix}1&0&-1\\2&1&0\end{bmatrix}\quad \text {and}\quad M_S=\begin{bmatrix}-1&2\\1&-2\end{bmatrix}\]
respectively. Describe the actions of \(T\), \(S\), and \(S\circ T\) geometrically, as in Example 2.
Complete the Explanation of Example 14 by verifying that \(T\circ S=\id _{\RR ^2}\).
Let \(T:\RR ^2\rightarrow \RR ^2\) be a linear transformation given by
\[T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}2x-5y\\-x+3y\end{bmatrix}\]
Propose a candidate for the inverse of \(T\) and verify your choice using Definition 13.
\[T^{-1}\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}\answer {3}x+\answer {5}y\\x+\answer {2}y\end{bmatrix}\]
Explain why linear transformation \(T:\RR ^2\rightarrow \RR ^2\) given by
\[T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}2x+2y\\-3x-3y\end{bmatrix}\]
does not have an inverse.
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 \(S\circ T\).