$f:M(n,\mathbb C) \to \mathbb C$ be a linear transformation such that $f(AB)=f(BA)$ , then $\exists k \in \mathbb C$ such that $f(A)=kTrace (A)$? Let $f:M(n,\mathbb C) \to \mathbb C$ be a linear transformation such that $f(AB)=f(BA) , \forall A,B \in M(n,\mathbb C)$ , then is it true that $\exists k \in \mathbb C$ such that $f(A)=kTrace (A),\forall A \in M(n,\mathbb C) $ ?
I can only figure out that $\exists X\in M(n,\mathbb C)$ such that  $f(A)=Trace (AX),\forall A \in M(n,\mathbb C) $ and that $f$ is continuous . 
Please help . Thanks in advance 
 A: If it's linear then the kernel contains the subspace generated by commutators. Everyone should compute $[e_{ij},e_{k\ell}]=e_{ij}e_{k\ell}-e_{k\ell}e_{ij}=\delta_{jk}e_{i\ell}-\delta_{\ell i}e_{kj}$ once in their life.
Thus the subspace includes $[e_{ij},e_{j\ell}]=e_{i\ell}$ when $e_{i\ell}$ is any off-diagonal basis matrix and it also includes $[e_{ij},e_{ji}]=e_{ii}-e_{jj}$. These are easily seen to span the entire subspace of traceless matrices, which is also the kernel of $\mathrm{tr}$. If two linear functionals have the same kernel, then they are equal up to a constant multiple, so $f(A)=k\,\mathrm{tr}(A)$ for some $k$ (with $k=0$ a special case).
A: You already know that there exists $X\in M(n,\mathbb C)$ such that $f(A) = {\rm Trace}(AX)$ for all $A$. Together with the fact that ${\rm Trace}(AB) = 0$ for all $A$ implies that $B=0$ (as follows by letting $A = B^*$), this is a short computation: For all $A,B \in M(n,\mathbb C)$ we have
$$
{\rm Trace}(ABX) = f(AB) = f(BA) = {\rm Trace}(BAX) = {\rm Trace}(AXB),
$$
and hence ${\rm Trace}\bigl(A\cdot (BX-XB)\bigr) = 0$ for all $A\in M(n,\mathbb C)$. Therefore, $BX-XB = 0$ for all $B\in M(n,\mathbb C)$, which shows that $X$ commutes with all matrices in $M(n,\mathbb C)$ and is thus of the form $X = \lambda\cdot E_n$ for some $\lambda\in \mathbb C$ ($E_n$ being the identity matrix). It follows that
$$
f(A) = {\rm Trace}(AX) = {\rm Trace}(\lambda A) = \lambda\cdot {\rm Trace}(A)
$$
for all $A\in M(n,\mathbb C)$.
A: Note, as you've stated, that $f(A) = Tr(AX)$ for some fixed $X$.  It suffices to find constraints on $X$.  Note that in the case in which $A = uv^T$ and $B = yz^T$, we have
$$
f(AB) = Tr(uv^Tyz^TX) = Tr(v^Tyz^TXu) = [v^Ty][z^TXu]\\
f(BA) = \cdots = [z^Tu][v^TXy]
$$
That is: for all vectors $u,v,y,z \in \Bbb C^n$, we have
$$
[v^Ty][z^TXu] = [z^Tu][v^TXy] \implies\\
\frac{z^TXu}{z^Tu} = \frac{v^TXy}{v^Ty}
$$
whenever the relevant denominators are non-zero.
That is, there exists a fixed $k \in \Bbb C$ such that $v^TXy = k(v^Ty)$.
If you're familiar with the notion of the bilinear form induced by a matrix, then this should enough to convince you that $X = kI$.  If not, take $v,y$ from the standard basis vectors, and see what you discover about $X$.
