Prove two complex matrices have null trace

Question

Let $A,B \in \mathbb{C}^{2 \times 2} \setminus \{O_2\}$, where $AB=-BA$ and $\det(A+B)=0$. Prove that $\operatorname{tr}(A) = \operatorname{tr}(B) = 0$ (where $\operatorname{tr}$ is the trace).

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My attempt

Let $C=A+B$. Then, using the Cayley–Hamilton for $C$:

$C^2 -Tr(C)C + \det(C)I_2=0$ therefore $C^2=Tr(C)C$ and $(A+B)^2=(Tr(A) + Tr(B))(A+B)$ then $$A^2 + B^2=(Tr(A) + Tr(B))(A+B) \tag1$$

Using Cayley–Hamilton for $A$ and $B$ in (1) one gets: $$ -\det(A)I_2 -\det(B)I_2=Tr(A)B + Tr(B)A \tag2$$

Applying trace to (2): $$-\det(A) -\det(B)=Tr(A)Tr(B) \tag3$$ So I have to prove $\det(A)=-\det(B)$ and here I got stuck.

Update 1

Showing $Tr(A)=Tr(B)$ is enough to finish the proof.

Update 2

Following an idea by @Dietrich Burde

Suppose $A$ is invertible. Then $B=A^{-1}(-B)A$ and $Tr(B)=Tr(A^{-1}(-B)A)=Tr(A^{-1}(-BA))=Tr((-BA)A^{-1})=Tr(-B)=-Tr(B)$.

Therefore $Tr(B)=0$. Now, from (3) one gets $\det(A)=-\det(B)$ then from (2) $Tr(A)=0$

Similar for $B$ invertible.

If both matrixes are singular then, from (3) one of $Tr(A)$ or $Tr(B)$ is zero and from (2) the other one is also zero.

Answer 1

Since the matrices are $2\times 2$, a direct computation is also successful. This is not the perfect solution, but shows that the claim is correct. Let $$ A=\begin{pmatrix} a_1 & a_3 \\ a_2 & a_4 \end{pmatrix},\quad B=\begin{pmatrix} b_1 & b_3 \\ b_2 & b_4 \end{pmatrix}. $$ Then $AB+BA=0$ and $\det(A+B)=0$ are equivalent to the system of polynomial equations $$ 2a_1b_1 + a_2b_3 + a_3b_2=0 \\ a_1b_3 + a_3b_1 + a_3b_4 + a_4b_3=0 \\ a_1b_2 + a_2b_1 + a_2b_4 + a_4b_2=0 \\ a_2b_3 + a_3b_2 + 2a_4b_4=0 \\ a_1a_4 + a_1b_4 - a_2a_3 - a_2b_3 - a_3b_2 + a_4b_1 + b_1b_4 - b_2b_3=0 $$ Assume first that $a_1=0$. Then Buchberger's algorithm gives either $B=0$ or $A=0$ or $a_4=a_1=0$, $b_1+b_4=0$, i.e., $tr(A)=tr(B)=0$ and $b_4^2 + b_2b_3 + a_2a_3=0$.

In the second case, we may assume that $a_1=1$. Then Buchberger's algorithm immediately gives $1+a_4=0, b_1+b_4=0$, or $B=0$. This finishes the proof.

Something more elegant for a special case: Suppose that $A,B$ are invertible. Then $AB=-BA$ means that $B=A^{-1}(-B)A$, so that $B$ and $-B$ are similar. In particular we obtain $tr(B)=0$. Similarly $tr(A)=0$.

Answer 2

Not sure what's the trick, but the following should work.

Suppose the contrary that $\operatorname{tr}(A)\ne0$. As $A$ and $B$ anticommute, we have $AAB = -ABA = BAA$, i.e. $A^2$ commutes with $B$. Yet, $A^2=\operatorname{tr}(A)A - \det(A)I$. So, $A$ also commutes with $B$. Hence $AB=BA=0$ and both $A,B$ are singular (because they are nonzero by assumption).

Let $C=A+B$. From Cayley-Hamilton theorem, we get $C^2=\operatorname{tr}(C)C-\det(C)I$. Apply $\det(C)=0$ and $AB=BA=0$, we obtain $A^2+B^2=(\operatorname{tr}(A)+\operatorname{tr}(B))(A+B)$. Apply Cayley-Hamilton theorem again, and use the fact that $A$ and $B$ are singular, we further obtain $\operatorname{tr}(A)B+\operatorname{tr}(B)A=0$. Since $B\ne0$ and $\operatorname{tr}(A)\ne0$, we conclude that $B$ is a nonzero multiple of $A$. But then $AB=0$ implies that $A$ is nilpotent, which is impossible because $\operatorname{tr}(A)\ne0$. Thus $A$ must be traceless. Similarly $B$ is traceless too.