<s>[INST] Change of basis matrix for polynomial Given the transformation $T : \mathbb{P}_3 \rightarrow \mathbb{P}_4$ defined as $T(\mathbf{p}(t)) = (-t)\mathbf{p}(t)$
where $\mathbf{p}(t) = t^3 + 6t - 2$
I am first supposed to find $T$, which should be easy enough and gives: $$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t$$
Now, what I can't do is to find the matrix $M$ relative to the bases $B = \{1, t, t^2, t^3\}$ and $C = \{1, t, t^2, t^3, t^4\}$
Looking in my book and other threads here, It seemed like the appropriate way is to insert every element of $B$ into $T$ and set up a matrix with the coefficients.
$$T(\mathbf{p}(1)) = -(1)^4 - 6(1)^2 + 2(1) = -5$$
$$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t $$
$$T(\mathbf{p}(t^2)) = -(t^2)^4 - 6(t^2)^2 + 2(t^2) = -t^8 - 6t^4 + 2t^2$$
$$T(\mathbf{p}(t^3)) = -(t^3)^4 - 6(t^3)^2 + 2(t^3) = -t^{12} - 6t^6 + 2t^3$$
which gives
\begin{equation}
M = 
\begin{bmatrix}
-5 & 0 & 0 & 0\\
 0 & 2 & 0 & 0\\
 0 & -6 & 2 & 0\\
 0 & 0 & 0 & 2\\
 0 & -1 & -6 & 0
\end{bmatrix}
\end{equation}
Happy with any helps or hints, thanks
 [/INST] Since $T(t^n)=-t^{n+1}$ i would say it is:
\begin{equation}
M = 
\begin{bmatrix}
-1 & 0 & 0 & 0\\
 0 & -1 & 0 & 0\\
 0 & 0 & -1 & 0\\
 0 & 0 & 0 & -1\\
 0 & 0 & 0 & 0
\end{bmatrix}
\end{equation}
Then
\begin{equation}
\begin{bmatrix}
-1 & 0 & 0 & 0\\
 0 & -1 & 0 & 0\\
 0 & 0 & -1 & 0\\
 0 & 0 & 0 & -1\\
 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
 0 \\
 6 \\
 -2 
\end{bmatrix}=
\begin{bmatrix}
-1 \\
 0 \\
 -6 \\
 2\\
0 
\end{bmatrix}
\end{equation}
</s>