Problem: Let $A$ and $B$ be real orthogonal matrices, $n$x$n$, where $n$ is an odd number. Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular.
What have I done so far:
-Since matrices $A$ and $B$ are real orthogonal, it means that their determinants are $-1$ or $+1$
First I observed matrix $A+B$: $$A+B=AI+BI=ABB^T+BAA^T=AB(B^T+A^T)=AB(A+B)^T\Rightarrow$$
So, if $detA=detB$ we have $det(A+B)=det(A+B)$ which tells me nothing. But, if $detA=-detB$ we have $det(A+B)=-det(A+B)$ which can only hapen if $det(A+B)=0$ making $A+B$ singular.
Then, I tried the same for $A-B$: $$A-B=AI-BI=ABB^T-BAA^T=AB(B^T-A^T)=AB(A-B)^T\Rightarrow$$
So, if $detA=-detB$ we have $det(A-B)=-det(A-B)$ which can happen if $det(A-B)=0$ making $A-B$ singular.
Is this correct or should I do matrix $A-B$ differently? I am also a little confused why does it have to be said that $n$ is an odd number? What should that tell me?
Any help is greatly appreciated.