Let's say I have a linear transformation L: $\Bbb R [x] _{\le 2}$ to $\Bbb R [x] _{\le 2}$, a base B (the canonical for simplicity) and that I have the matrix $\Bbb[L]_{\ B}$ . I know how to calculate the diagonal of L, but I have no idea on how to calculate this:
$\ L^n(1-x^2)$
Can somebody help, please?
Suppose that $[L]_B$ is a matrix which represents a linear transformation $L:R[x]_{\le 2}\rightarrow R[x]_{\le 2}$ with respect to the standard basis of polynomials.
Let $A$ be the change of basis matrix from the standard basis to the basis which the which the matrix of the linear transformation $L$ is a diagonal matrix $D$.
Then $[L]_B =A^{-1}DA$ and $[L^n]_B=A^{-1}D^nA$
If $$ D=\begin{bmatrix} a&0&0\\ 0&b&0\\ 0&0&c\\ \end{bmatrix}$$
then $$D^n=\begin{bmatrix} a^n&0&0\\ 0&b^n&0\\ 0&0&c^n\\ \end{bmatrix}$$
Using this, you can find the coordinates of $L^n(1-x^2)$ with respect to the standard basis by multiplying the coordinates of $1-x^2$ in the standard basis by the matrix $A^{-1}D^nA$