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.

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
$$
-2E_1+E_2+2E_4=
-2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
+\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+2\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}
=\begin{bmatrix} 7 & 2 \\ -2 & -3\end{bmatrix}
$$
