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. 
 A: First,we know that $T$ is a linear transfromation.
Then,we just need to considering a basis of $M_{2\times 2}$.There we choose a basis as following.
$$\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right),\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right),\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right),\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
We observe that 
$$\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
$$\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)$$
so,$T\left(\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)\right)=\mathbf{0}$.
similarly,$T\left(\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 0\\\end{array}\right)\right)=\mathbf{0}$.
One can observe that 
$$\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\left(\begin{array}{c}0 & 1\\1& 0\\\end{array}\right)=\left(\begin{array}{c}0 & 1\\1& 0\\\end{array}\right)\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)$$
Hence,$$T\left(\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\right)=T\left(\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\right)=\dfrac{1}{2}T\left(\left(\begin{array}{c}1 & 0\\0& 1\\\end{array}\right)\right)$$.
we conclude that $$T(A)=T\left(\left(\begin{array}{c}a & b\\c& d\\\end{array}\right)\right)=aT\left(\left(\begin{array}{c}1 & 0\\0& 0\\\end{array}\right)\right)+bT\left(\left(\begin{array}{c}0 & 1\\0& 0\\\end{array}\right)\right)+cT\left(\left(\begin{array}{c}0 & 0\\1& 0\\\end{array}\right)\right)+dT\left(\left(\begin{array}{c}0 & 0\\0& 1\\\end{array}\right)\right).$$
$$=\dfrac{1}{2}tr(A)T(I_2)$$
The proof is completed.
A: 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
$$
T(A)=T(X)+T(Y)=T(Y)=\frac12\operatorname{tr}(A)T(I_2).
$$
