0
$\begingroup$

Let $T$$:$ $P2$$P2$ be a linear transformation such that $T(p)(x)$ = $e^{-x}\left(\frac{d^2}{dx^2}\left(e^xp\left(x\right)\right)\right)$.

a) Find the matrix of $T$ relative to the basis $\left\{x^2,\:x,\:1\right\}$.

b) Find the matrix of $T$ relative to the basis $\left\{x^2-x,\:2x+1,\:x-1\right\}$.

c) Does there exist a basis $B$ of $P2$ such that $[T]B$ is diagonal?

For a) and b), I simply plugged in the values of the basis as the input to $p$.

So for a), $T(p)(x^2)$ $=$ $x^2+4x+2$$T(p)(x)$ $=$ $x+2$ and $T(p)(1)$ $=$ $1$.

So I got the matrix:

$\begin{pmatrix}1&0&0\\ 4&1&0\\ 2&2&1\end{pmatrix}$

For b), I got that $T(p)(x^2-x)$ $=$ $x^2+3x$, $T(p)(2x+1)$ $=$ $2x+5$ and $T(p)(x-1)$ $=$ $x+1$.

So I got the matrix:

$\begin{pmatrix}1&0&0\\ 3&2&1\\ 0&5&1\end{pmatrix}$

For c), I thought it was false since the matrix for the standard basis itself is not diagonal. And I can't seem to pinpoint any other scenario in which only one element of each degree is present.

So if anyone can tell if what I have done is right or not and, if not, what the right approach would be, I would be very grateful!

$\endgroup$
1
$\begingroup$

a) It looks correct.

b) Note that $T(x^2-x)=x^2+3x=(x^2-x)+\frac43(2x+1)+\frac43(x-1)$. So, the entries of the first column of the matrix should be $1$, $\frac43$, and again $\frac43$.

c) Your argument is not correct. However, if such a basis existed, then the entries of the main diagonal of $T(P)_B$ would all have to be equal to $1$, since $T(P)_B$ would be a diagonal matrix similar to the one that you got in the first answer. But the only matrix similar to the identity matrix is the identity matrix itself.

$\endgroup$
  • $\begingroup$ Thank you for your answer. So just to confirm, the only problem with part b) is the first column? So if I change that then the matrix will be correct? Or do I have to change the other columns as well? Thank you! $\endgroup$ – Dev SR Nov 3 '18 at 23:07
  • $\begingroup$ You will have to change the other colunns as well. $\endgroup$ – José Carlos Santos Nov 3 '18 at 23:08
  • $\begingroup$ Ok, I see. This is basically like solving a system of equations. Thanks for your help! $\endgroup$ – Dev SR Nov 3 '18 at 23:11

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.