If $A^2+B^2=0_n$ and $AB-BA$ is invertible, then $n$ is even.

Question

Let $A,B$ be $n\times n$ matrices. Suppose that $A^2+B^2=0_n$ and $AB-BA$ is invertible. Show that $n$ is even

My approach to this problem is that:
$A^2 + B^2 = 0_n\implies A^2 = -B^2\implies \det(A^2) = (-1)^n \det(B^2)$ and if $A$ and $B$ is invertible then $n$ has to be even since $\det(A^2) = (\det A)^2$
However, I am unable to prove $A$ and $B$ is invertible and I still could not use the information $AB-BA$ is invertible

Please give me the answer to this problem, thank you.

Answer 1

Notice that $$(A+iB)(A-iB)=A^2+i(BA-AB)+B^2=i(BA-AB)$$ where clearly $A-iB=\overline{A+iB}$. Now by Binet and the properties of conjugation we have that $$\det[(A+iB)(\overline{A+iB})]=\det(A+iB)\ \overline{\det(A+iB)}\geq 0$$ is a real number, and $$\det[i(BA-AB)]=i^n \det(BA-AB)\neq 0$$ since $BA-AB$ is invertible. Thus $n$ must be even (otherwise this is not a real number).

By the way this problem is basically the same as Problem 3 from the IMC 1997.

Answer 2

There's a flavor of Jacobson's Lemma here and the implication that $n$ is even holds in any field where Newton's Identities have maxmal effect-- i.e. any field of characteristic zero or when $n\lt \text{char } \mathbb F $.
(Trivial note: OP's problem statement cannot exist when $n=1$ since the commutator would be zero.)

Let $C:=AB-BA$ be the commutator. And consider

$C\mapsto AC + CA $
$= A\big(AB-BA\big) + \big(AB-BA\big)A=A^2B-ABA +ABA -BA^2=\big(-B^2\big)B-B\big(-B^2\big) = \mathbf 0$
$\implies AC=-CA$
$\implies AC^2 = (AC)C=(-CA)C= -C(AC)=-C(-CA)=C^2A$
and by induction $A$ commutes with all even powers of $C$. Now examine the trace:

for odd $r\in \mathbb N$
$\text{trace}\Big(C^r\Big)$
$= \text{trace}\Big( C^{r-1} C\Big)$
$= \text{trace}\Big( C^{r-1}\big({AB} - {BA}\big)\Big) $
$= \text{trace}\Big(C^{r-1}{AB} - C^{r-1}{BA}\Big) $
$= \text{trace}\Big(C^{r-1}{AB}\Big) - \text{trace}\Big(C^{r-1}{BA}\Big) $
$= \text{trace}\Big(AC^{r-1}B\Big) - \text{trace}\Big(C^{r-1}{BA}\Big) $
$= \text{trace}\Big(C^{r-1}BA\Big) - \text{trace}\Big(C^{r-1}{BA}\Big) $
$=0$

Finally, write out the characteristic polynomial for $C$
$p(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} +... + a_{1}x^{1}+ a_0$
where $a_n =1$ since we've defined the characteristic polynomial to be monic and $a_{n-1} =0 = -\text{trace}\big(C\big)$
which is a Base Case for the inductive hypothesis on $r$, that $a_{n-r } = 0$ for positive odd $r\leq n$. Applying Newton's Identities:

$a_{n-r} = -\frac{1}{r} \sum_{j=1}^{r} a_{n-r + j}\cdot \text{trace}\big(C^j\big) =0\text{ when r is odd}$
since $\text{trace}\big(C^j\big)=0$ for odd $j$ (proven above) and $a_{n-r + j}=0$ when $j$ is even (strong induction hypothesis)

For the case of $r =n$, we know $\det\big(C\big)\neq 0\implies a_{n-n}=a_0\neq 0\implies n\text{ is even}$