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Let $A,B$, two matrices with the order of $n\times n$.
Given that $AB + BA = 0$ and $A,B$ are invertible (meaning, there are $A^{-1}, B^{-1}$).
Prove that $n$ must be even number.

$$\eqalign{ & AB + BA = 0 \cr & AB = - BA \cr & \left| {AB} \right| = \left| { - BA} \right| \cr & \left| {AB} \right| = {( - 1)^n}\left| {BA} \right| \cr} $$

Assuming $n$ is odd, then:
$$\left| {AB} \right| = - \left| {BA} \right|$$

I don't see something wrong with the last equation.
Maybe I'm missing something, or it's not the right way..

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2 Answers 2

up vote 5 down vote accepted

$$\det (AB)=\det(BA)=\det A\det B$$

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Oh I totally forgot about it. Thanks! – AnnieOK Apr 10 '14 at 15:13

You get $ABA^{-1} = -B$, so $\det B = \det(ABA^{-1}) = \det(-B) = (-1)^n \det(B)$. So if $\det B \neq 0$ what can you say about $n$?

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