A linear transformation such that $T(AB)=T(BA)$ The question goes as follows:
Let $V$ be a vector space and let $T: M_{2 \times 2} (R) —> V$ such that $T(AB)=T(BA)$ for all $A, B \in M_{2 \times 2}$.  Show that $T(A) = 1/2(trA)T(I2)$ for all $A \in M_{2 \times 2}$. 
I have no clue how to approach this. I’ve tried everything but I keep going in circles. Please help me. 

Note that if $X\in M_2(\mathbb R)$ has a zero trace, it must lie inside the linear span of $\{AB-BA:A,B\in M_2(\mathbb R)\}$:
\begin{aligned}
X=\pmatrix{a&b\\ c&-a}
&=\frac12\left[\pmatrix{0&1\\ 1&0}\pmatrix{0&-a\\ a&0}-\pmatrix{0&-a\\ a&0}\pmatrix{0&1\\ 1&0}\right]\\
&+\frac12\left[\pmatrix{1&0\\ 0&-1}\pmatrix{0&b\\ -c&0}-\pmatrix{0&b\\ -c&0}\pmatrix{1&0\\ 0&-1}\right].
\end{aligned}
(In fact, it can be shown that the set $\{AB-BA:A,B\in M_2(\mathbb R)\}$ itself --- without linear span --- is precisely the set of all real $2\times2$ traceless matrices, but this is unimportant here.) Therefore $T(X)=0$ whenever $\operatorname{tr}(X)=0$. Now, for every matrix $A\in M_2(\mathbb R)$, we can split it into the sum of a traceless component $X=A-\frac12\operatorname{tr}(A)I_2$ and a trace component $Y=\frac12\operatorname{tr}(A)I_2$. It follows that
$$
T(A)=T(X)+T(Y)=T(Y)=\frac12\operatorname{tr}(A)T(I_2).
$$
