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!


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.

  • $\begingroup$ 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

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