How to find the matrix of a linear transformation from $\Bbb R^{2\times2}$ to $P_2$ Let $T:\Bbb R^{2\times2}\rightarrow P_2$ be a linear transformation, where $\Bbb R^{2\times2}$ is the vector space of all $2\times2$ matrices and $P_2$ is the vector space of all polynomials up to second order.
We consider the basis $B$ for $R^{2×2}$,
$$B = \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix},\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix},{\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix},\begin{bmatrix}0 & 0\\
0 & 1\end{bmatrix}}, $$
and the basis $C$ for the vector space $P_2$,
$$\ C = \begin{pmatrix}
1 & t & t^{2}
\end{pmatrix}.$$
The transformation $T$ is defined by
$$T\begin{pmatrix}
a & b  \\ c &d\end{pmatrix} = (c + d) + (b −c)t + (2a + b)t^{2}.$$
How do you find the matrix representation of $T$ relative to the chosen bases $B$ and $C$?
I think the first step would be to take the transformation $T$ of each basis vector $B_1 = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$, $B_2=\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$, $B_3=\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}$, $B_4 = \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}$. If that's correct, what's the intuitive idea behind this?  Why would I want to see how the basis vectors transform?
 A: To find a matrix for a linear transformation $T: V \rightarrow W$, you just need to apply T to the basis of V. i.e.  Assume $\{e_1, e_2, ..., e_n\}$ is the basis of $V$, then the matrix of T consists of $T(e_1), T(e_2), ..., T(e_n)$ as its columns (with the same order of the basis).
In your case, $ e_1=\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix},
e_2=\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},e_3=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}, e_4=\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}.$ Now if you calculate $T(e_i)$ you get a 3 by 4 matrix. For example $T(e_1)=2t^2= (0,0,2)$ so the first column of $T$ is $\begin{pmatrix}
0\\
0 \\
 2
\end{pmatrix}.$
A: You need to find out how basis vectors transform under $T$:
$$
\begin{split} 
T\begin{pmatrix} 
1 & 0 \\ 
0 & 0 
\end{pmatrix} &= 2t^2 \\
T\begin{pmatrix}
0 & 1 \\
0 & 0 
\end{pmatrix} &= t + t^2 \\
T\begin{pmatrix}
0 & 0 \\
1 & 0 
\end{pmatrix} &= 1-t \\
T\begin{pmatrix}
0 & 0 \\
0 & 1 
\end{pmatrix} &= 1.
\end{split}
$$
Then, the columns of the matrix of the transformation, are the transformed basis vectors, i.e.
$$T=\begin{pmatrix}
0 & 0 & 1 & 1\\
0 & 1 &-1 & 0\\
2 & 1 & 0 & 0
\end{pmatrix}.
$$
