Subsection 1.4.3 Matrix associated to a composition
Suppose that \(U, V,\) and \(W\) are vector spaces over a field \(F\text{,}\) and \(S:U\to V\) and \(T:V\to W\) are linear maps. The the composition \(T\circ S\) (usually denoted \(TS\)) is a linear map, \(T\circ S: U\to W.\)
Now suppose that all three vector spaces are finite-dimensional, say \(\dim U = n,\) \(\dim V = p,\) and \(\dim W = m\text{,}\) with bases \(\cB_U, \cB_V, \cB_W.\) If we consider the matrices of the corresponding linear maps, we see that the matrix sizes are
\begin{gather*}
[S]_{\cB_U}^{\cB_V} \text{ is } p\times n\\
[T]_{\cB_V}^{\cB_W} \text{ is } m\times p\\
[TS]_{\cB_U}^{\cB_W} \text{ is } m\times n
\end{gather*}
The fundamental result connecting these is
Theorem 1.4.4. Matrix of a composition.
\begin{equation}
[TS]_{\cB_U}^{\cB_W} = [T]_{\cB_V}^{\cB_W}[S]_{\cB_U}^{\cB_V} \tag{1.4.4}
\end{equation}
This result will be of critical importance when we discuss change of basis.
As more or less a special case of the above theorem, we have the corresponding result with coordinate vectors: that the coordinate vector of \(T(v)\) is the product of the matrix of \(T\) with the coordinate vector of \(v\text{.}\) More precisely,
Corollary 1.4.5.
With the notation as above, for \(v \in V\)
\begin{equation*}
[T(v)]_{\cB_W} = [T]_{\cB_V}^{\cB_W} [v]_{\cB_V}.
\end{equation*}
Example 1.4.6.
Let \(V=P_4(\R)\) and \(W=P_3(\R)\) be the vector spaces of polynomials with coefficients in \(\R\) having degree less than or equal to 4 and 3 respectively. Let \(D:V\to W\) be the (linear) derivative map, \(D(f) =
f'\text{,}\) where \(f'\) is the usual derivative for polynomials. Let’s take standard bases for \(V\) and \(W,\) namely \(\cB_V=\{1, x, x^2, x^3, x^4\}\) and \(\cB_W=\{1, x, x^2, x^3\}.\) One computes:
\begin{equation*}
[D]_{\cB_V}^{\cB_W}=
\begin{bmatrix}
0&1&0&0&0\\
0&0&2&0&0\\
0&0&0&3&0\\
0&0&0&0&4\\
\end{bmatrix}
\end{equation*}
Let \(f=2+3x+5x^3.\) We know of course that \(D(f) =
3+15x^2,\) but we want to see this with coordinate vectors. We know that
\begin{equation*}
[f]_{\cB_V} = \begin{bmatrix}2\\3\\0\\5\\0\end{bmatrix}
\text{ and }
[D(f)]_{\cB_W} = \begin{bmatrix}3\\0\\15\\0\end{bmatrix}
\end{equation*}
and verify that
\begin{equation*}
[D(f)]_{\cB_W} = \begin{bmatrix}3\\0\\15\\0\end{bmatrix}=
[D]_{\cB_V}^{\cB_W}[f]_{\cB_V}=
\begin{bmatrix}
0&1&0&0&0\\
0&0&2&0&0\\
0&0&0&3&0\\
0&0&0&0&4\\
\end{bmatrix}\begin{bmatrix}2\\3\\0\\5\\0\end{bmatrix}.
\end{equation*}
Subsection 1.4.4 Change of basis
A change of basis or change of coordinates is an enormously useful concept. It plays a pivotal role in diagonalization, triangularization, and more generally in putting a matrix into a canonical form. It’s practical uses are easy to envision. We may think of the usual orthonormal basis of \(\R^3\) along the coordinate axes as the standard basis for \(\R^3\text{,}\) but when one want to create computer graphics which projects the image of an object onto a plane, the natural frame includes a direction parallel to the line of sight of the observer, so it defines a natural basis for this application.
First, let’s understand what we are doing intuitively. Suppose our vector space \(V = \R^3\text{,}\) and we have two bases for it with elements written as row vectors, \(\cB_1=\{e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)\}\) and \(\cB_2=\{v_1=(1,1,1), v_2=(0,1,1), v_3=(0,0,1)\}.\)
Checkpoint 1.4.7. Is \(\cB_2\) really a basis?
Let’s recall a useful fact that allows us to quickly verify that \(\cB_2\) is actually a basis for \(\R^3.\) While in principle we must check the set is both linearly independent and spans \(\R^3\text{,}\) since we know the dimension of \(\R^3\text{,}\) and the set has 3 elements, it follows that either condition implies the other.
Hint.
To show \(\cB_2\) spans, it is enough to show that \(\Span(\cB_2)\) contains a spanning set for \(\R^3\)
Normally when we think of a vector in \(\R^3\text{,}\) we think of it as a coordinate vector with respect to the standard basis, so that a vector we write as \(v=(a,b,c)\) is really the coordinate vector with respect to the standard basis:
\begin{equation*}
v=[v]_{\cB_1} = \begin{bmatrix}a\\b\\c\\\end{bmatrix}
\end{equation*}
The problem is when we want to find \([v]_{\cB_2}.\) For some vectors this is easy. For example,
\begin{equation*}
[v]_{\cB_1} =\begin{bmatrix}1\\2\\3\end{bmatrix} \text{
is equivalent to } [v]_{\cB_2} = \begin{bmatrix}1\\1\\1\end{bmatrix},
\end{equation*}
or
\begin{equation*}
[v]_{\cB_1} =\begin{bmatrix}1\\3\\6\end{bmatrix} \text{
is equivalent to } [v]_{\cB_2} = \begin{bmatrix}1\\2\\3\end{bmatrix},
\end{equation*}
but what is going on in general?
Recall from
Corollary 1.4.5, that for a linear transformation
\(T:V\to W\text{,}\) and
\(v \in V\) that
\begin{equation*}
[T(v)]_{\cB_W} = [T]_{\cB_V}^{\cB_W} [v]_{\cB_V}.
\end{equation*}
In our current situation \(V=W\) and \(T\) is the identity transformation, \(T(v) = v\text{,}\) which we shall denote by \(I\text{,}\) so that
\begin{equation*}
[v]_{\cB_2} = [I]_{\cB_1}^{\cB_2} [v]_{\cB_1}.
\end{equation*}
The matrix \([I]_{\cB_1}^{\cB_2}\) is called the change of basis or change of coordinates matrix (converting \(\cB_1\) coordinates to \(\cB_2\) coordinates), and these change of basis matrices come in pairs
\begin{equation*}
[I]_{\cB_1}^{\cB_2} \text{ and } [I]_{\cB_2}^{\cB_1}.
\end{equation*}
Now in our case, both matrices are easy to compute:
\begin{equation*}
[I]_{\cB_1}^{\cB_2}=
\ba{rrr}
1&0&0\\
-1&1&0\\
0&-1&1\\
\ea
\text{ and } [I]_{\cB_2}^{\cB_1}=
\begin{bmatrix}
1&0&0\\
1&1&0\\
1&1&1\\
\end{bmatrix},
\end{equation*}
and it should come as no surprise that the columns of the second are just the elements of the \(\cB_2\)-basis in standard coordinates. But the nice part is that the first matrix is related to the second affording a means to compute it when computations by hand are not so simple.
Using Equation
(1.4.4) on the matrix of a composition
\begin{equation*}
[TS]_{\cB_U}^{\cB_W} =
[T]_{\cB_V}^{\cB_W}[S]_{\cB_U}^{\cB_V},
\end{equation*}
with \(V=U=W\text{,}\) and \(T=S=I\text{,}\) we arrive at
\begin{equation*}
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{bmatrix}=
[I]_{\cB_1}^{\cB_1} =
[I]_{\cB_1}^{\cB_2}[I]_{\cB_2}^{\cB_1},
\end{equation*}
that is \([I]_{\cB_1}^{\cB_2}\) and \([I]_{\cB_2}^{\cB_1}\) are inverse matrices, and this is always the case.
Theorem 1.4.8.
Given two bases \(\cB_1\) and \(\cB_2\) for a finite-dimensional vector space \(V\text{,}\) the change of basis matrices \([I]_{\cB_1}^{\cB_2}\text{ and
}[I]_{\cB_2}^{\cB_1}\) are inverse matrices.
Finally we apply this to the matrix of a linear map \(T:V\to
V\) on a finite-dimensional vector space \(V\) with bases \(\cB_1\) and \(\cB_2\text{:}\)
Theorem 1.4.9.
\begin{equation*}
[T]_{\cB_2} =
[I]_{\cB_1}^{\cB_2}[T]_{\cB_1}[I]_{\cB_2}^{\cB_1}.
\end{equation*}
Example 1.4.10.
We often express the matrix of a linear map in terms of the standard basis, but many times such a matrix is complicated and does not easily reveal what the linear map is actually doing. For example, using our bases \(\cB_1\) and \(\cB_2\) for \(\R^3\) given above, suppose we have a linear map \(T:\R^3 \to \R^3\) whose matrix with respect to the standard basis \(\cB_1\) is
\begin{equation*}
[T]_{\cB_1}=\ba{rrr}
4&0&0\\
-1&5&0\\
-1&-1&6\\
\ea.
\end{equation*}
It is easy enough to compute the value of
\(T\) on a given vector (recall from equation
(1.4.3), the columns of the above matrix are simply
\(T(v_1), T(v_2),T(v_3)\) written with respect to the standard basis (
\(\cB_1\)) for
\(\R^3).\)
\begin{equation*}
[T]_{\cB_2}=\begin{bmatrix}
4&0&0\\
0&5&0\\
0&0&6\\
\end{bmatrix},
\end{equation*}
which makes much clearer how the map \(T\) is acting on \(\R^3\) (strecthing by a factor of 4, 5, 6 in the directions of \(w_1, w_2, w_3.\)
