If $A$ and $ B$ be real invertible invertible matrices such that $AB=-BA$ then Trace Let $A$ and $B$ be real invertible matrices such that $AB = - BA$. Then 
1.$Trace(A)=Trace(B)=0$
2.$Trace(A)=Trace(B)=1$
3.$Trace(A)=0,Trace(B)=1$
4.$Trace(A)=1,Trace(B)=0$
$A$ is invertible $\Rightarrow$ $ABA^{-1}= -B$$\Rightarrow$ $B$ and $-B$ are similar. Hence $B$ and $-B$ have same eigenvalues which is possible only if $Trace(B)=0$ but still i am not getting proper reason to say that $Trace(B)=0$?

$$2\mathrm{trace}(B) = \sum_{\lambda \mathrm{\ eigenv \ of \ } B} \lambda + \sum_{\lambda \mathrm{\ eigenv \ of \ } -B} \lambda = \sum_{\lambda \mathrm{\ eigenv \ of \ } B} \lambda+\sum_{\lambda \mathrm{\ eigenv \ of \ } B} -\lambda = 0$$
That's because $B$ and $-B$ have opposite eigenvalues.
