Anti-commutative matrices

If $A$ and $B$ are anti-commutative square matrices, so $AB+BA=0$, how do you

a) prove that $\mathrm{tr}(A)=\mathrm{tr}(B)=0$ and
b) prove that the order of the matrices is even?

Solution 1:

If the matrices are non-singular, then writing $A=-BAB^{-1}$ and taking the trace, we get $\mathrm{tr}A=-\mathrm{tr}A$. Hence $\mathrm{tr}A=0$, and the procedure for $B$ is analogous.

Next compute the determinant of both sides of $AB=-BA$: this yields $\mathrm{det}\,A\,\mathrm{det}\,B=(-1)^N\mathrm{det}\,B\,\mathrm{det}\,A$, where $N$ stands for size of matrices. Now since the $A,B$ are non-singular, both sides of the equality are non-zero and the equality is possible only for even $N$.