Linear maps: Matrices of Linear Maps
Coordinate transformations
Our rules for coordinate transformations allow us to switch to other bases. This yields a trivial but important feature:
Basis transition
If #\alpha # and #\beta# are bases for a finite-dimensional vector space #V#, and #L :V\rightarrow V# is a linear map, then the matrices #L_\beta# and #L_\alpha# are related to each other by the equality \[L_\beta ={}_\beta I_\alpha \, L_\alpha \, {}_\alpha I_\beta\]
In the second example of Matrix of a linear map we saw that \[D_\alpha =\left(\,\begin{array}{rrr}
0 & 1 & 0\\
0 & 0 & 2\\
0 & 0 & 0
\end{array}\,\right)\] is the matrix of differentiation #D: P_2\to P_2# on the vector space #P_2# of all polynomials of degree at most #2# with respect to the basis #\alpha=\basis{1,x,x^2}#. We now take a different basis, #\beta=\basis{x^2-x,x^2+3,x^2-1}# for #P_2# and try to find the matrix #D_\beta# of #D#. Doing this directly is not so easy, but we can use:
\[
D_\beta={}_\beta I_\alpha \, D_\alpha \, {}_\alpha I_\beta
\] Comparing #\alpha# to #\beta# shows without any calculations that
\[
{}_\alpha I_\beta =\left(\,\begin{array}{rrr}
0 & 3 & -\,1 \\
-\,1 & 0 & 0\\
1 & 1 & 1
\end{array}\,\right)
\] so
\[
{}_\beta I_\alpha = {}_\alpha I_\beta^{-1}=\frac14\left(\,\begin{array}{rrr}
0 & -\, 4 & 0\\
1 & 1 & 1\\
-\,1 & 3 & 3
\end{array}\,\right)
\] giving
\[
D_\beta={}_\beta I_\alpha \, D_\alpha \, {}_\alpha I_\beta=\frac14\left(\,\begin{array}{rrr}
-\,8 & -\, 8 & -\,8\\
1 & 2 & 2\\
7 & 6 & 6
\end{array}\,\right)
\] Again for illustration: #p(x)=2x^2-3x+5=3(x^2-x)+(x^2+3)-2(x^2-1)# has #\beta#-coordinates #\rv{3,1,-\,2}#, and
\[
D_\beta\ \left(\,\begin{array}{r}
3 \\ 1\\ -2\end{array}\,\right) = \frac14\ \left(\,\begin{array}{rrr}
-\,8 & -\,8 & -\,8\\
1 & 2 & 2\\
7 & 6 & 6
\end{array}\,\right) \, \left(\,\begin{array}{r}
3\\ 1\\ -\,2 \end{array}\,\right) = \frac{1}{4}\left(\begin{array}{c}
-16\\ 1 \\ 15 \end{array}\right)
\] These should be the #\beta#-coordinates of the derivative of #p(x)#, which checks because \[-\,4(x^2-x)+\frac14(x^2+3)+\frac{15}{4}(x^2-1)=4x-3=p'(x)\]
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