determinant of the product of two orthogonal matrices Let $A,B \in O_{n}$ where $n$ is odd.
Show that
\begin{equation}
\det((A+B)(A-B)) = 0
\end{equation}
I started with some basic rules for determinants:
\begin{align}
&\det((A+B)(A-B)) &= 0 \\
\iff &\det((-1) \cdot(A+B)(A+B)) &= 0 \\
\iff &(-1)^n\cdot2^n\cdot \det(A+B) &= 0
\end{align}
So when exactly is $\det(A+B)=0$ ?
 A: Observe that
$$\det((A+B)(A-B))=\det(A+B)\det(A-B)=\det(A)\det(I+C)\det(A)\det(I-C)$$
where $C=A^{-1}B$. As $A$ is orthogonal $\det(A)^2=1$. So
$$\det((A+B)(A-B))=\det(I+C)\det(I-C).$$
The matrix $C$ is orthogonal (why?) so we need to prove that one of
$\det(I\pm C)$ is zero when $C$ is orthogonal. Why might that be so?
(Remember we haven't used the oddness of $n$ yet.)
A: I am confused a lil bit with that $O_n$; I don't know what it is! Can I take that as $M_n$ (usual notation for the set of $n\times n$ matrices).
Let's then take $A$ to be $I_{3\times 3}$ and $B$ any ${3\times 3}$ matrix whose eigenvalues are (say) 2, 3 and 4. Then the statement asked to be proved is never actually going to be true. If I am wrong somehow, correct me please.
A: Transforming  
$\det((A+B)(A-B))=  \det(A+B)\det(A-B)=  \det(A+B)^T\det(A-B)= \det((A+B)^T(A-B))=  \det(A^TA-A^TB+ B^TA-B^TB)=  \det(I-A^TB+B^TA-I)=  \det(B^TA- A^TB)=  \det(B^TA- (B^TA)^T) = \det(C- C^T)$  
we get a determinant of some skew symmetric matrix $S=  C- C^T $.
The  determinant  of $S$ satisfies
$\det( S ) = \det( S^T ) = \det(− S) = (−1)^n\det(S)$ and with $n$ odd it means that $\det( S ) =- \det( S )$.  
Hence finally  $\det(S)=0 $.
