If $AB = -BA$, then tr$(A) = $ tr$(B) = 0 $ If $A, B$ are invertible matrices such that $AB = -BA$ ( which implies tr$(AB) =$ tr$(-BA)$), then show that tr$(A) = $ tr$(B) = 0$.
My attempt: I could only show that tr$(AB) = 0$.
 A: Use invertibility and invariance of the trace, indeed the eigenvalues, under conjugation (change of basis): 
$$
A=-BAB^{-1}
$$
so 
$$
\text{tr}(A)=-\text{tr}(BAB^{-1})=-\text{tr}(A)\implies \text{tr}(A)=0
$$
and identically for $B$.
A: Your surmise is correct: it is not generally true that both traces of $A$ and $B$ are $0$.
For real matrices $A,B$, ${\mathrm tr}(AB)$ is an inner product on $n$-by-$n$ matrices, so saying that is $0$ is merely an orthogonality relation. It does not imply orthogonality to the identity matrix, in general! :)
For example, ${\mathrm tr}\pmatrix{1&0\cr 0&1}\pmatrix{0&1\cr 1&0}=0$, meaning that these two matrices are orthogonal to each other with respect to that inner product, but certainly the first is not orthogonal to the identity (since it is the identity).
EDIT: note that if the trace of both $AB$ and $BA$ are $0$, then the one is the negative of the other.
EDIT-EDIT: I haven't carefully kept track, but the other answers seem to be answering a slightly different question, namely, with the assumption that $AB=-BA$. But the question as I've read it only assumes something about traces. ?
EDIT-EDIT-EDIT: the question title and body do not match...
A: Since the matrices are invertible, we note (by multiplying on the right by $B^{-1}$) that $A = -BAB^{-1}$ and hence
$$
\operatorname{tr}(A) = -\operatorname{tr}(BAB^{-1})
$$
However, since we also have $\operatorname{tr}(A) = \operatorname{tr}(BAB^{-1})$, we conclude that $\operatorname{tr}(A) = 0$.
Symmetrically, conclude that $B$ has trace $0$.
