For $2 \times 2$ matrices $AB=-BA$ with $BA$ not $0$, prove that $\mathrm{tr}(A)=\mathrm{tr}(B)=\mathrm{tr}(AB)=0$

Question

It is easy to derive from $AB=-BA$ that $\mathrm{tr}(AB)=0$ since $\mathrm{tr}(AB)=\mathrm{tr}(-BA)=-\mathrm{tr}(BA)=-\mathrm{tr}(AB)$. However, I cannot get that $\mathrm{tr}(A)=\mathrm{tr}(B)=0$ without the fact that $A$ and $B$ are invertible.

My Professor suggested I use the Cayley-Hamilton Theorem. However, that just gives me a few extra conditions on the elements of $A$ and $B$, and I still can't get that their traces equal $0$.

Any ideas are greatly appreciated!

Answer 1

The Cayley-Hamilton theorem says that $$ A^2-\operatorname{tr}(A)A+\det(A)I=0 $$ If we multiply by $B$ on the right: $$ A^2B-\operatorname{tr}(A)AB+\det(A)B=0 $$ If we multiply by $B$ on the left: $$ BA^2-\operatorname{tr}(A)BA+\det(A)B=0 $$ Subtracting the two relations: $$ A^2B-BA^2-2\operatorname{tr}(A)AB=0 $$ On the other hand, $A^2B=AAB=-ABA=BAA=BA^2$.

Answer 2

Here is an alternative proof without using Cayley-Hamilton theorem. We assume that the underlying field has characteristic $\ne2$, otherwise $A=B=\operatorname{diag}(1,0)$ would give a counterexample. As you said, $\operatorname{tr}(AB)=0$. It remains to show that $\operatorname{tr}(A)=\operatorname{tr}(B)=0$.

If both $A$ and $B$ are invertible, then $ABA^{-1}=-B$ and $B^{-1}AB=-A$. Taking traces on both sides of each equation, the conclusion follows.

If at least one of the two matrices is singular, we may assume without loss of generality that $B=uv^T$. The condition $AB=-BA\ne0$ implies that $Auv^T=-uv^TA\ne0$. Hence $\{Au,u\}$ and and $\{v^TA,v^T\}$ are two pairs of nonzero parallel vectors. Thus $Au=\lambda u$ for some $\lambda\ne0$. Substitute this into $Auv^T=-uv^TA$, we also get $v^TA=-\lambda v^T$. Therefore, both $\lambda$ and $-\lambda$ are eigenvalues of $A$. By assumption, the field has characteristic $\ne2$. Hence $\lambda$ and $-\lambda$ are distinct eigenvalues of $A$ and $\operatorname{tr}(A)=0$.

Finally, as $A$ has distinct nonzero eigenvalues, it is invertible. So, our previous argument ($ABA^{-1}=-B$) shows that $\operatorname{tr}(B)=0$. Alternatively, the conclusion also follows from the observation that $0=\operatorname{tr}(AB)=\operatorname{tr}(Auv^T)=\operatorname{tr}(\lambda uv^T)=\lambda\operatorname{tr}(B)$.