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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?

I could start with implication:

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

But I am not sure if this implication is correct.

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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)?

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  • $\begingroup$ Very nice. ${}{}{}$ $\endgroup$ – copper.hat Sep 12 '13 at 5:32
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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.

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  • 2
    $\begingroup$ In which book are such proofs available. Can somebody suggest a textbook for matrices $\endgroup$ – Maverick Jan 18 '16 at 14:05

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