Matrix for polynomial linear map

Question

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?

Answer 1

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 $$ \begin{cases} T(1) &= 2 \cdot 2 + 0 = 2,\\ T(1 + x) &= 2 (1 + x) + 1,\\ T(x^{2}) &= 2 x^{2} + 2 x = 2 x^{2} + 2 (1 + x) - 2 \cdot 1. \end{cases}$$

Answer 2

Yes, it is correct.

You cound also solve the problem without a change-of -basis matrix. Just compute $T(1)$; it is $2$. Now, which are the coordinates of $1$ in the given basis? Since $2=2\times1+0\times(x+1)+0\times x^2$, they are $2$, $0$, and $0$. So, now you have the first column of your matrix. You'll get the second and the third columns if you do the same thing with $x+1$ and with $x^2$ respectively.