Q:

Matrix for polynomial linear map

I am looking to represent $T(f) = 2f + f'$ as a matrix $B$ with respect to the basis $\{1, 1+x, x^2\}$. I am used to this type of question only when the basis involves vectors in $\mathbb{R}^n$. My guess was to think of the basis $\{1, x, x^2\}$ in terms of the standard $\mathbb{R}^3$ basis.
So the 'change of basis' matrix would be
$C = \begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$
and I found that $T(1) = 2(1)$; $T(x) = 1(1) + 2(x)$; $T(x^2) = 2(x) + 2(x^2)$. So according to my earlier guess, the matrix for T with respect to the basis $\{1, x, x^2\}$ is
$A = \begin{bmatrix}2&1&0\\0&2&2\\0&0&2\end{bmatrix}$
Which gives me the answer
$B = C^{-1}AC = \begin{bmatrix}2&5&2\\0&2&2\\0&0&2\end{bmatrix}$
Is this correct and do I have the right/wrong idea?

A:

I do not think you have taken the conjugate correctly, as you should obtain
$$
B 
= 
C^{-1}AC 
= 
\begin{bmatrix}
2 & 1 & -2\\
0 & 2 & 2\\
0 & 0& 2
\end{bmatrix}.
$$
Let us check
$$
\begin{cases}
T(1) &= 2 \cdot 2 + 0 = 2,\\
T(1 + x) &= 2 (1 + x) + 1,\\
T(x^{2}) &= 2 x^{2} + 2 x = 2 x^{2} + 2 (1 + x) - 2 \cdot 1.
\end{cases}$$

