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.
Let
\(U\),
\(V\) 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
\[(S\circ T)(\vec {u})=S(T(\vec {u}))\]
Define
\[T:\RR ^2\rightarrow \RR ^2 \quad \text {by}\quad T\left (\begin{bmatrix}u_1\\u_2\end{bmatrix}\right )=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}\]
\[S:\RR ^2\rightarrow \RR ^2 \quad \text {by}\quad S\left (\begin{bmatrix}v_1\\v_2\end{bmatrix}\right )=\begin{bmatrix}3v_1-v_2\\-3v_1+v_2\end{bmatrix}\]
(You should be able to verify that both transformations are linear.) Examine the
effect of
\(S\circ T\) on vectors of
\(\RR ^2\).
From the computational standpoint, the situation is simple.
\begin{align*} (S\circ T)\left (\begin{bmatrix}u_1\\u_2\end{bmatrix}\right )&=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 )\\ &=\begin{bmatrix}3(u_1+u_2)-(3u_1+3u_2)\\-3(u_1+u_2)+(3u_1+3u_2)\end{bmatrix}\\ &=\vec {0} \end{align*}
This means that \(S\circ T\) maps all vectors of \(\RR ^2\) to \(\vec {0}\).
In addition to the computational approach, it is also useful to visualize what happens
geometrically.
First, observe that
\[T\left (\begin{bmatrix}u_1\\u_2\end{bmatrix}\right )=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}=(u_1+u_2)\begin{bmatrix}1\\3\end{bmatrix}\]
Therefore the image of any vector of
\(\RR ^2\) under
\(T\) lies on the line determined by the
vector
\(\begin{bmatrix}1\\3\end{bmatrix}\).
Even though \(S\) is defined on all of \(\RR ^2\), we are only interested in the action of \(S\) on vectors
along the line determined by \(\begin{bmatrix}1\\3\end{bmatrix}\). Our computations showed that all such vectors map to
\(\vec {0}\).
The actions of individual transformations, as well as the composite transformation
are shown below.
The composition of two linear transformations is linear.
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*}
Composition of linear transformations is associative. In other words, for linear
transformations
\(T\),
\(S\) and
\(R\)
We have \((R\circ S)\circ T=R\circ (S\circ T)\).
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.
Let
\(T:\RR ^n\rightarrow \RR ^m\) and
\(S:\RR ^m\rightarrow \RR ^p\) be linear transformations with standard matrices
\(M_T\) and
\(M_S\), respectively. Then
the composite transformation
\(S\circ T:\RR ^n\rightarrow \RR ^p\) has a standard matrix given by the product
\(M_SM_T\).
For all \(\vec {v}\) in \(\RR ^n\) we have:
\[(S\circ T)(\vec {v})=S(T(\vec {v}))=S(M_T\vec {v})=M_S(M_T\vec {v})=(M_SM_T)\vec {v}\]
In Example
2, we discussed a composite transformation
\(S\circ T:\RR ^2\rightarrow \RR ^2\) given by:
\[T\left (\begin{bmatrix}u_1\\u_2\end{bmatrix}\right )=\begin{bmatrix}u_1+u_2\\3u_1+3u_2\end{bmatrix}\quad \text {and} \quad S\left (\begin{bmatrix}v_1\\v_2\end{bmatrix}\right )=\begin{bmatrix}3v_1-v_2\\-3v_1+v_2\end{bmatrix}\]
Express
\(S\circ T\) as a matrix transformation.
The standard matrix for
\(T:\RR ^2\rightarrow \RR ^2\) is
\[\begin{bmatrix}1&1\\3&3\end{bmatrix}\]
and the standard matrix for
\(S:\RR ^2\rightarrow \RR ^2\) is
\[\begin{bmatrix}3&-1\\-3&1\end{bmatrix}\]
The standard matrix for
\(S\circ T\) is the product
\[\begin{bmatrix}3&-1\\-3&1\end{bmatrix}\begin{bmatrix}1&1\\3&3\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\]
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.
Associativity of Matrix Multiplication Let
\(A\),
\(B\) and
\(C\) be matrices of appropriate
dimensions so that the product
\((AB)C\) is defined. Then
\[(AB)C=A(BC)\]
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
\(S\).
Let
\(V\) and
\(W\) be vector spaces, and let
\(T:V\rightarrow W\) be a linear transformation. A transformation
\(S:W\rightarrow V\)
that satisfies
\(S\circ T=\id _V\) and
\(T\circ S=\id _W\) is called an
inverse of
\(T\). If
\(T\) has an inverse,
\(T\) is called
invertible.
Let
\(T:\RR ^2\rightarrow \RR ^2\) be a transformation defined by
\(T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}x+y\\x-y\end{bmatrix}\). (How would you verify that
\(T\) is linear?) Show
that
\(S:\RR ^2\rightarrow \RR ^2\) given by
\(S\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}0.5x+0.5y\\0.5x-0.5y\end{bmatrix}\) is an inverse of
\(T\).
We will show that
\(S\circ T=\id _{\RR ^2}\).
\begin{align*} (S\circ T)\left (\begin{bmatrix}x\\y\end{bmatrix}\right )&=S\left (T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )\right )=S\left (\begin{bmatrix}x+y\\x-y\end{bmatrix}\right )\\ &=\begin{bmatrix}0.5(x+y)+0.5(x-y)\\0.5(x+y)-0.5(x-y)\end{bmatrix} =\begin{bmatrix}x\\y\end{bmatrix} \end{align*}
We leave it to the reader to verify that \(T\circ S=\id _{\RR ^2}\).
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.
Suppose
\(T:V\rightarrow W\) is an invertible linear transformation. Let
\(S\) be an inverse of
\(T\). Then
\(S\) is linear.
The proof of this result is left to the reader. (See Practice Problem ??)
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 \(n=m\).
Let
\(T:\RR ^n\rightarrow \RR ^n\) be a linear transformation, and let
\(M\) be the standard matrix of
\(T\).
-
(a)
- (Existence of Inverses.) \(T\) is invertible if and only if \(M\) is invertible. If \(T\) is
invertible, then the inverse is induced by \(M^{-1}\).
-
(b)
- (Uniqueness of Inverses.) If \(S\) is an inverse of \(T\), then \(S\) is unique.
Part (a) follows directly from the preceding discussion. Part (b) 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
\(S\circ T\).
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