find the matrix representation of $T$ relative to $B,B'$ where $T(p(x)) = (x+1)p'(x^2)$ Hi I really need help with this linear algebra question.
Let $T:P_2\to P_3$ be defined by $T(p(x)) = (x+1)p'(x^2)$ and let $B = (1,x+1,x^2+x)$ and $B' = (x^3,x^3+x,x^2+x,x+1)$ be ordered basis for $P_2$ and $P_3$ respectively.
a.) find the matrix representation of $T$ relative to $B,B'$
b.) Use your answer for part a.) to compute $T(x^2+3x+5)$
So far I have evaluated:
$T(p(1)) = 0$
$T(p(x+1)) = 2x^2+4x+2$
$T(p(x^2+x)) = 4x^4+10x^3+8x^2+2x$
but this is where i get confused, the third polynomial has degree 4 which is not in $P_3$ right? did i make an error in my calculation somewhere or am i going about this completely wrong? any help is greatly appreciated thanks.
 A: So, first of all, the expression $T(p(1))$ doesn't make sense. The symbol $p(x)$ is supposed to represent a polynomial. So your computation would look like this:
If $p(x) = 1$, then $p'(x) = 0$, and so $p'(x^2) = 0$. So $T(1) = (x+1)(0) = 0$.
If $p(x) = x + 1$, then $p'(x) = 1$, and so $p'(x^2) = 1$. So $T(x + 1) = (x + 1)(1) = x + 1$.
If $p(x) = x^2 + x$, then $p'(x) = 2x + 1$, and so $p'(x^2) = 2x^2 + 1$. So $T(x^2 + x) = (x + 1)(2x^2 + 1) = 2x^3 + 2x^2 + x + 1$.
Now you have to find each of your answers in $B'$ coordinates.
$T(1) = 0$, which is $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ in $B'$ coordinates.
$T(x + 1) = x + 1$, which is the fourth element of $B'$. So $T(x + 1) = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$.
$T(x^2 + x) = 2x^3 + 2x^2 + x + 1$. It isn't obvious how to write this as a linear combination of elements in $B'$, but we can figure this out using 'standard' coordinates on $P_3$. That is, using coordinates from the basis $\{x^3,x^2,x,1\}$. The resulting linear system is:
$$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}\mathbf{x} = \begin{bmatrix} 2 \\ 2 \\ 1 \\ 1 \end{bmatrix}$$
Where $\mathbf{x}$ is the desired $B'$-coordinate vector of $2x^3 + 2x^2 + x + 1$. Some row reduction gives:
$\mathbf{x} = \begin{bmatrix} 4 \\ -2 \\ 2 \\ 1 \end{bmatrix}$
And so the matrix for $T$ is:
$$\begin{bmatrix} 0 & 0 & 4 \\ 0 & 0 & -2 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix}$$
