Finding a formula for a linear transformation Let $T: M_{22} \rightarrow\mathbb{R}^2$ be a linear transformation such that:
$$T\left(\begin{bmatrix}  1& 2\\3&4\end{bmatrix}\right) \rightarrow(5,5)$$
$$T\left(\begin{bmatrix}  1& 0\\0&1\end{bmatrix}\right) \rightarrow(2,0)$$
$$T\left(\begin{bmatrix}  0& 1\\1&0\end{bmatrix}\right) \rightarrow(0,2)$$
$$T\left(\begin{bmatrix}  4& 3\\2&2\end{bmatrix}\right) \rightarrow(6,5)$$
Give a formula for $T$ what is the kernel of $T$?
To be honest, I'm not completely sure where to start this one. Any hints to send me in the right direction would be vastly helpful. Thank you.
 A: You might find some clever trick as suggested in comments, but there is a standard way for describing the kernel.
Set
$$
E_1=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix},\quad
E_2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad
E_3=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\quad
E_4=\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix},\quad
$$
You should have already proved that $\mathscr{B}=\{E_1,E_2,E_3,E_4\}$ is a basis for $M_{22}$. The matrix of the map $T$ with respect to this basis and the standard basis on $\mathbb{R}^2$ is
$$
A=\begin{bmatrix}
5 & 2 & 0 & 6 \\
5 & 0 & 2 & 5
\end{bmatrix}
$$
and an easy reduction to echelon form gives
$$
\begin{bmatrix}
1 & 0 & 2/5 & 1 \\
0 & 1 & -1 & 1/2
\end{bmatrix}
$$
A basis for the null space of $A$ is thus
$$
\left\{
  v_1=\begin{bmatrix}-2\\5\\5\\0\end{bmatrix},
  v_2=\begin{bmatrix}-2\\1\\0\\2\end{bmatrix}
\right\}
$$
This means that the kernel of $T$ is generated by
$$
-2E_1+5E_2+5E_3=
-2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
+5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+5\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
=\begin{bmatrix}3 & 1 \\ -1 & -3\end{bmatrix}
$$
and
$$
-2E_1+E_2+2E_4=
-2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
+\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+2\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}
=\begin{bmatrix} 7 & 2 \\ -2 & -3\end{bmatrix}
$$
