I have this problem :

$A,B \in M_n(\Bbb R)$ matrices while $n$ odd number.

Proof if $AB+BA=0$ then atleast one of $A,B$ is singular.

I assume that $A,B$ invertible.

Its clear that $\det(AB) \neq 0$ and $\det(BA) \neq 0$.

If I could show that, $\det(AB) \neq -\det(BA)$ then I can conclude that $\det(AB)+\det(BA) \neq 0$.

But I don't seem to find a way to show that, I guess it has something to do with $n$ odd number.

Any ideas? Thanks!

  • $\begingroup$ $\det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA)$ for all $A,B$. This is because multiplication of real numbers is commutative. $\endgroup$ Dec 3, 2014 at 4:25
  • $\begingroup$ You seem to have a sign problem: $x\neq y$ does not imply $x+y\neq 0$. $\endgroup$ Dec 3, 2014 at 10:56
  • $\begingroup$ @MarcvanLeeuwen Yes, your right I forget the $-$, Edited. $\endgroup$
    – JaVaPG
    Dec 3, 2014 at 12:29

4 Answers 4


Recall that the determinant is completely multiplicative over square matrices, i.e., $\det(AB) = \det(A) \det(B)$. We have \begin{align} AB & = -BA\\ \det(AB) & = -\det(BA)\\ \det(A) \det(B) & = - \det(B) \det(A)\\ \det(A) \det(B) & = 0 \end{align}

  • 4
    $\begingroup$ Note: we have $\det(-M) = -\det(M)$ because $n$ is odd. In general, $\det(kM) = k^n M$ for an arbitrary scalar $k$. $\endgroup$ Dec 3, 2014 at 3:22
  • $\begingroup$ @Omnomnomnom Yes, true. $\endgroup$
    – Adhvaitha
    Dec 3, 2014 at 3:23
  • $\begingroup$ @Adhvaitha Can you explain why $\det(A) \det(B) = - \det(B) \det(A)\implies \det(A) \det(B) = 0$ $\endgroup$
    – JaVaPG
    Dec 3, 2014 at 3:29
  • 1
    $\begingroup$ @JaVaPG: Calling $a-\det(A)$ and $b=\det(B)$ (these are just numbers, so multiplication commutes for them) one has $ab=-ba\implies2ab=0\implies ab=0$ since $2\neq0$. $\endgroup$ Dec 3, 2014 at 11:09
  • 1
    $\begingroup$ @JaVaPG no. This equality holds for every $n$. The fact that $n$ is odd is important since $\det(-AB) = -\det(AB)$ because $n$ is odd. $\endgroup$ Dec 3, 2014 at 18:45

If $C$ is an $n\times n$ matrix with $n$ odd, then $$\det(-C)=-\det(C)\ .$$ So for your equation, $$\eqalign{ AB+BA=O &\Rightarrow AB=-BA\cr &\Rightarrow \det(A)\det(B)=-\det(B)\det(A)\cr &\Rightarrow \det(A)\det(B)=0\cr &\Rightarrow \det(A)=0\ \hbox{or}\ \det(B)=0\ .\cr}$$

  • $\begingroup$ Can you explain how you manage to conclude $-\det(B)det(A)=0 \implies det(A)det(B)=0$ $\endgroup$
    – JaVaPG
    Dec 3, 2014 at 3:46
  • $\begingroup$ @JaVaPG: both $\det(A)$ and $\det(B)$ are numbers, hence they commutative. $\endgroup$
    – Mher
    Dec 4, 2014 at 11:32

here is a proof that does not use the determinant explicitly.

If $B$ is singular, we are done. Next suppose $B$ is nonsingular. Claim: $A$ is singular. this proves that at least one of $A$ or $B$ is singular.

Proof of the claim: $AB = -BA.$ so $A = B(-A)B^{-1},$ that is $A$ and $-A$ are similar. that means the eigenvalues must come in pairs $(\lambda, -\lambda)$ if $\lambda \neq 0.$ because the order is odd, at least one of the eigenvalues of $A$ must be zero. that proves the claim $A$ is singular.


Determinants are just numbers and a = -a implies a = 0 for a,any number.


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