### 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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