Prove that $T(K) = a\ \mathrm{trace}(K)$ if $T(XY)=T(YX)$ Let $T: M_{n,n}(\Bbb R) \to \Bbb R$ be a linear transformation such that $T(XY)=T(YX)$ for any two $n\times n$ matrices $X,\,Y$. How do you prove that $T(K)=a\ \mathrm{trace}(K)$ for some scalar $a$ in $\Bbb R$ for every $K$ in $M_{n,n}(\Bbb R)$?
I know that $\mathrm{trace}(XY)=\mathrm{trace}(X) + \mathrm{trace}(Y)$. My guess is that $\mathrm{trace}(XY)$ must be proportional to $T(XY)$, but I don't know how to state this mathematically. 
 A: Hint: Consider when $X$ and $Y$ are matrices of the form $E_{i,j}$ which is the matrix with a one in the $(i, j)$-entry, but zeroes elsewhere. Find out what $E_{i,j}E_{k, l}$ is (there are some possibilities depending on whether the indices are distinct and so on). From this you should be able to derive $T(E_{i,j})$ for every $i, j$ which will allow you to conclude that $T(K) = a\operatorname{trace}(K)$ for some $a \in \mathbb{R}$.
A: You need an extra condition on $T$ to ensure that $T$ is (a scalar multiple of) the trace :
$$
T(\text{diag}(x,0,0,\ldots,0)) = x
$$
This, together with @Michael Albanese's answer should do the trick.
A: Although probably not what you want, here is a proof from a higher perspective. $T(XY)=T(YX)$ implies that $T(XY-YX)=0$. Since the set of all matrix commutators over $\mathbb{R}$ is exactly equal to the set of all traceless matrices, the given property implies that
$$
T(A)=T
\ \underbrace{\left(A-\frac{\mathrm{trace}(A)}{n}I\right)}_{\text{traceless}}
+\frac{\mathrm{trace}(A)}{n}T(I)=\frac{\mathrm{trace}(A)}{n}T(I)
$$
and the result follows with $a=\frac{T(I)}{n}$.
