If $\mathbf{A}$ and $\mathbf{B}$ are real orthogonal matrices of the same order and $|\mathbf{B}|+|\mathbf{A}|=\mathbf{0}$ Prove that $|\mathbf{A}+\mathbf{B}|=\mathbf{0}$
My Approach:-
$|\mathrm{A}|+|\mathrm{B}|=0$
$\Rightarrow|A|=-|B|$
$|\mathrm{A}| \cdot|\mathrm{B}|=-1 \quad\left[\because|\mathrm{B}|=\left|B^{-1}\right| \text {as they are orthogonal }\right]$
Let, $C=A\left(A^{T}+B^{T}\right) B$
$\Rightarrow|C|=\left|A A^{T} B+A B^{T} B\right|=|B+A| \ldots \ldots \ldots$ (i)
And $|\mathrm{C}|=|\mathrm{A}|\left|A^{T}+B^{T}\right||B|=-\left|A^{T}+B^{T}\right|$
$\Rightarrow-\left|(A+B)^{T}\right|=-|A+B| \ldots \ldots \ldots \ldots$ (ii)
$|A+B|=-|A+B|$
$\Rightarrow 2|A+B|=0$
$\boxed{\Rightarrow|A+B|=0}$
I am looking for another short approach.Any alternate solution would be greatly appreciated
$\def\norm[1]{\lVert #1 \rVert}$Here is a proof that I think gives more intuition. Let $X$ be a real orthogonal matrix.
Lemma 1. $|\lambda(X)| = 1$
Proof. $Xu=\lambda u \Rightarrow |\lambda|^2 \lVert u \rVert^2 = \lVert Xu \rVert^2 = u^* X^TXu = \lVert u \rVert^2$
Lemma 2. If $|X|=-1$ then $X$ has an odd number of $-1$ eigenvalue.
Proof. If $\lambda$ is complex then $\bar{\lambda}$ is also an eigenvalue of $X$. So, $-1=|X|=\prod_i \lambda_i = (-1)^k$ since multiplication of other eigenvalues results in $1$.
Answer. $|A|+|B|=0 \Rightarrow |A^TB| = -1$, which means $A^TB$ (which is also orthogonal) has at least 1 eigenvalue of $-1$, so $|I+A^TB|=0$ and consequently $|A+B|=0$.