This is a homework problem I thought figured out, but the homework site says it's not correct.

The linear transformation $T_n : P_n \rightarrow P_n $ defined as $T(p(x)) = \frac{\partial}{\partial x}(x \cdot p(x))$

Eg. $T_0(1) = 1$ , $T_1(x) = 2x$ , $T_2(x^2) = 3x^2$

Find the matrix which represents $T_2$ with respect to the basis B = {$b_1, b_2, b_3$} = \begin{pmatrix}1-x \\-x^2-x\\ -x^2-x-1 \end{pmatrix}

So what I did was to write $T$(old basis) in terms of new basis:

$T_0(1) = 1 = b_2 - b_3$

$T_1(x) = 2x = -2b_1 + 2b_2 - 2b_3$

$T_2(x^2) = 3x^2 = 3b_1 - 6b_2 +3b_3$

which gives the matrix

\begin{pmatrix}0&-2&3 \\1&2&-6\\ -1&-2&3 \end{pmatrix}

The homework site does not agree with my answer, but I can't see what I did wrong. The matrix seems to work for every vector I test it against. Did I do something wrong? Or is the homework site wrong?


That's because what you have to do is to compute $T(b_1)$, $T(b_2)$, and $T(b_3)$ with respect to the basis $B$, not $T(e_1)$, $T(e_2)$, and $T(e_3)$, where $e_1=1$, $e_2=x$ and $e_3=x^2$.

So, since, for instance, $T(b_1)=1-2x=2b_1-b_2+b_3$, the entries of the first column of your matrix will be $2$, $-1$, and $1$.


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