Trace identity of product of matrices I have to express the trace of a product of matrices A and B in terms of traces of individual matrices. I know that $\operatorname{tr}(AB) \neq \operatorname{tr}(A)\operatorname{tr}(B)$, and for the tensor product $\operatorname{tr}(A \otimes B) = \operatorname{tr}(A) \operatorname{tr}(B)$. However, is there a nice function $f(\cdot)$ such that $\operatorname{tr}(AB) = f(\operatorname{tr}(A), \operatorname{tr}(B))$? 
Any help will be greatly appreciated.
Regards,
H.A.S.
 A: For $n \times n$ matrices, $n \geq 2$, there is not:
Taking $A = B = {\mathbf 0}$ gives $$\text{tr } A = \text{tr } B = \text{tr} (AB) = 0 ,$$ so if there were such a function, it would satisfy $f({\mathbf 0}, {\mathbf 0}) = 0$. On the other hand, if $n = 2$, taking $A = B = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$ gives $\text{tr } A = \text{tr } B = 0$ but $$\text{tr}(AB) = \text{tr} \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = 2,$$ so $f({\mathbf 0}, {\mathbf 0}) = 2$, a contradiction.
For $n > 2$, taking a direct sum $A = B = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} \oplus {\mathbf 0}$ leads to the same contradiction.
For $1 \times 1$ matrices (as voldemort points out), we do indeed have $\text{tr}(AB)=\text{tr } A \cdot \text{tr } B$---in fact, the trace in this dimension coincides with the usual ring isomorphism of the space of $1 \times 1$ matrices with the space of scalars---so in that dimension $f(x, y) = xy$ satisfies the condition.
A: You already know that $tr(AB) \neq tr(A)tr(B)$- this fact alone gives you that such an $f$ can't exist.
For instance, suppose that such an $f$ exists. Then considering $1$ by $1$ matrices (or in other words, just complex numbers), we get:
$xy=f(x,y)$. So, existence of this function implies $tr(AB)=tr(A)tr(B)$ for all matrices $A$, $B$- which is false.
A: I don’t follow. The mere fact that in dimension 1 we have a product doesn’t imply that a general function would be just a product. Maybe there is a function that in one dimension is just multiplication but in other dimensions is something else.
Granted the intuition is that such a function Durant exist. But a proof of non existence must be more robust than this. The following argument is not sound:

*

*For dimension one tr(AB)=tr(A)tr(B)

*For a dimension rather than 1, i.e 2 the formula above fails
Hence no formula Can exist.

Without having any assumption on  the supposed function it’s not possible to prove it can’t exist.
