# Is sum of two orthogonal matrices singular?

I am trying to solve following problem.

Let $$A, B \in \mathbb{R}^{n\times n}$$ be an orthogonal matrices and $$\det(A) = -\det(B)$$. How can it be proven that $$A+B$$ is singular?

$$\det(A)=-\det(B) \Rightarrow B$$ is created from $$A$$ by swapping two lines or columns.

But I am not sure if this implication is correct.

Since $A$ is invertible and $A+B = A(I+A^{-1}B)$, it suffices to show that $I + A^{-1}B$ is singular.
So, observe that $$\det(A^{-1}B) = \det(A)^{-1}\det(B) = \det(A)^{-1}(-\det(A)) = -1,$$ and recall, in general, that if $S$ is a complex matrix, then $\det(S)$ is the product of all the eigenvalues of $S$, counted with algebraic multiplicity. What does this imply, then, about the set of eigenvalues of the real orthogonal matrix $A^{-1}B$ (as a complex unitary matrix)?
• Very nice. ${}{}{}$ – copper.hat Sep 12 '13 at 5:32
As $A$ and $B$ are real orthogonal and $\det(A)=-\det(B)$, we have $\det(A)\det(B)=-1$. Hence $$\det(A+B)=\det\left(A(B^T+A^T)B\right)=-\det(B^T+A^T)=-\det\left((B+A)^T\right)=-\det(B+A)$$ and the assertion follows.