Proof if $AB+BA=0$ Then atleast one of the matrices are singular. 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!
 A: 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}
A: If $C$ is an $n\times n$ matrix with $n$ odd, then
$$\det(-C)=-\det(C)\ .$$
So for your equation,
$$\eqalign{
  AB+BA=O
  &\Rightarrow AB=-BA\cr
  &\Rightarrow \det(A)\det(B)=-\det(B)\det(A)\cr
  &\Rightarrow \det(A)\det(B)=0\cr
  &\Rightarrow \det(A)=0\ \hbox{or}\ \det(B)=0\ .\cr}$$
A: 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.  
A: Determinants are just numbers and a = -a implies a = 0 for a,any number.
