If $A$ and $ B$ be real invertible invertible matrices such that $AB=-BA$ then Trace

Question

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 $Trace(B)=0$?

Answer 1

See $Tr(-B)$=$-Tr(B)$, $Tr(ABA^{-1})$=$Tr(B)$ So you have $Tr(B)=Tr(-B)=-Tr(B)$ and hence the result. Similarly $Tr(A)=0$.

Answer 2

What's non proper about your argument? In can be shortened a bit, though: since $B$ and $-B$ are similar, they have the same traces. Therefore $-\operatorname{tr}(B)=\operatorname{tr}(-B)=\operatorname{tr}(B)$ and so $\operatorname{tr}(B)=0$. By the same argument, $\operatorname{tr}(A)=0$.

Answer 3

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

Answer 4

You don't need to use eigen values because you don't know if they have real eigen values and using complex eigen values seems overkill.

For example with $$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ and $$B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ You have $BA = -AB$ with $A$ and $B$ invertible but $A$ doesn't have any real eigen values.

However, since $$\forall M,N \in M_n(\mathbb{R}), tr(MN) = tr(NM)$$ and $B = -ABA^{-1}$ $$tr(B) = tr(-ABA^{-1}) = -tr(ABA^{-1})= - tr(BA^{-1}A) = -tr(B)$$ therefore $tr(B) = 0$

Similarly $tr(A) = 0$