Subsection 4.3.3 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 4.3.7. Matrix of a composition.
\begin{equation}
[TS]_{\cB_U}^{\cB_W} = [T]_{\cB_V}^{\cB_W}[S]_{\cB_U}^{\cB_V} \tag{4.3.4}
\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 4.3.8.
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 4.3.9.
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 4.3.4 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. Its 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 4.3.10. 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 4.3.8, 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
(4.3.4) 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 4.3.11.
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 4.3.12.
\begin{equation*}
[T]_{\cB_2} =
[I]_{\cB_1}^{\cB_2}[T]_{\cB_1}[I]_{\cB_2}^{\cB_1}.
\end{equation*}
Example 4.3.13. A simple example.
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
(4.3.3), 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 \(w_1, w_2, w_3.\)
We return to the example of the orthogonal projection from above and show how we computed the matrix of the transformation with respect to the standard basis.
Example 4.3.14. The details from our other orthogonal projection.
Recall that we wanted to define a map from \(T:\R^3\to \R^3\) which geometrically takes a point in three space with coordinates \((x,y,z)\) and projects orthogonally onto the plane \(x+y+z=0.\)
We constructed a basis \(\cB = \{v_1, v_2, v_3\}\) with \(v_1,
v_2\) in the plane and orthogonal to each other, and the third vector \(v_3\) orthogonal to the plane. We chose
\begin{equation*}
\cB=\left\{v_1=\ba{r}1\\-1\\0\ea,\ v_2=\ba{r}1\\1\\-2\ea, \
v_3=\ba{r}1\\1\\1\ea\right\}.
\end{equation*}
The matrix of \(T\) with respect to the basis \(\cB\) which is the natural basis for this problem is
\begin{equation*}
[T]_\cB = \ba{rrr}1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0 \amp
0\ea,
\end{equation*}
just as with the standard orthogonal projection onto the \(xy\)-plane.
So to deduce \([T]_\cE\) from \([T]_\cB\text{,}\) we need to compute the change of basis matrices \([I]_\cB^\cE\) and \([I]_\cE^\cB\text{.}\) The matrix \([I]_\cB^\cE\) is the easy one since we are just listing the new basis \(\{v_1, v_2, v_3\}\) as its columns, so
\begin{equation*}
[I]_\cB^\cE = \ba{rrr}1\amp1\amp1\\-1\amp1\amp1\\0\amp -2\amp 1\ea.
\end{equation*}
To compute the other we have find the inverse of \([I]_\cB^\cE\) which we can do by row reducing the augmented matrix \(\left[ [I]_\cB^\cE \mid I_3\right]\text{.}\) We obtain:
\begin{equation*}
\ba{rrrrrr}1 \amp 1 \amp 1 \amp 1 \amp 0 \amp 0 \\
-1 \amp 1 \amp 1 \amp 0 \amp 1 \amp 0 \\
0 \amp -2 \amp 1 \amp 0 \amp 0 \amp 1
\ea \mapsto
\ba{rrrrrr}1 \amp 0 \amp 0 \amp \frac{1}{2} \amp -\frac{1}{2} \amp 0 \\
0 \amp 1 \amp 0 \amp \frac{1}{6} \amp \frac{1}{6} \amp -\frac{1}{3} \\
0 \amp 0 \amp 1 \amp \frac{1}{3} \amp \frac{1}{3} \amp \frac{1}{3}\ea
\end{equation*}
Thus
\begin{equation*}
[I]_\cE^\cB = \ba{rrr}\frac{1}{2} \amp -\frac{1}{2} \amp 0 \\
\frac{1}{6} \amp \frac{1}{6} \amp -\frac{1}{3} \\
\frac{1}{3} \amp \frac{1}{3} \amp \frac{1}{3}\ea
\end{equation*}
We can now compute that
\begin{equation*}
[T]_\cE = [I]_\cB^\cE [T]_\cB [I]_\cE^\cB = \ba{rrr}2/3\amp -1/3\amp -1/3\\
-1/3\amp2/3\amp-1/3\\-1/3\amp-1/3\amp2/3\ea.
\end{equation*}
