I would really appretiate some help about the following transformation matrices. We have to write a tranformation matrix in basis $B = \{ 1 + x, x + x^2, x^2 \}$ with a polynomial $(Ap)(x) = (x^2 - 2)p(1) - x ~ p'(x)$.
I do know how to write it in basis $\{1,x,x^2\}$, but I don't seem to get a correct answer for the basis $B$.
Thanks for your help in advance!
On
Let $\text{can} = \{1, x, x^2\}$ be the standard basis for $\mathcal{P}_2(\Bbb R)$. You claim that you got the matrix $[A]_{\text{can}}$. We want $[A]_{\text{B}}$, if I understood you right. If $[I]_{B, \text{can}}$ is the matrix obtained putting the coordinates of $B$'s polynomials, in the $\text{can}$ basis, in columns, then we have: $$[A]_B = [I]_{B, \text{can}}^{-1} \cdot [A]_{\text{can}}\cdot [I]_{B, \text{can}}$$
Denote the three elements of $B$ by $p_1$, $p_2$, $p_3$. We have \begin{align*} Ap_1(x) &= (x^2 - 2)p_1(1) - xp_1'(x)\\ &= 2x^2 - 4 - x\\ &= -4(1+x) + 3x + 2x^2\\ &= -4(1+x) + 3(x+x^2) - x^2\\ &= -4p_1(x) + 3p_2(x) - 1p_3(x) \end{align*} I'm sure, you can do the other two along the same lines.