# Matrix relative to a Linear Transformation. Diagonal Matrix.

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!

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.