1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.