Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular

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 $A+B$: $$A+B=AI+BI=ABB^T+BAA^T=AB(B^T+A^T)=AB(A+B)^T\Rightarrow$$

$$det(A+B)=det(AB)det(A+B)^T=det(A)det(B)det(A+B)$$

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 $A-B$: $$A-B=AI-BI=ABB^T-BAA^T=AB(B^T-A^T)=AB(A-B)^T\Rightarrow$$

$$det(A-B)=det(AB)det(A-B)^T=det(A)det(B)det(A-B)$$

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.

Solution 1:

(1) If $X$ is real $n\times n$ skew symmetric matrix, i.e., $X+X^T=0$, and if $T$ means transpose, then $$ {\rm det}\ X={\rm det}\ -X^T={\rm det}\ (-I)X={\rm det}\ X (-1)^n $$ where $I$ is an identity. So if $n$ is odd then ${\rm det}\ X=0$

(2) Assume that $A,\ B$ are orthogonal Then since $AA^T=I=BB^T$, $$ (A+B)(A-B)^T= BA^T - AB^T $$

Note that if $Y=BA^T$, then $(A+B)(A-B)^T=Y-Y^T$ is skew symmetric. So ${\rm det}\ (A+B)(A-B)^T =0$. So at leat one of $A+B,\ A-B$ is singular.

Solution 2:

You can write $$ A\pm B=A(\mathbf 1 \pm A^T B)=A(\mathbf 1 \pm Q), \quad Q\textrm{ orthogonal} $$ Then $A\pm B $ is singular iff $\mathbf 1 \pm Q $ is singular. But $Q$ has always an eigenvector $v$ with eigenvalue $\pm 1$ when $n$ is odd, therefore $\mathbf 1 -Q $ or $\mathbf 1 + Q$ has to be singular.