determinant of the product of two orthogonal matrices

Question

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 $\det(A+B)=0$ ?

Answer 1

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.)

Answer 2

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.

Answer 3

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 $.