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Question: Assume that $A$ is a $n\times n$ orthogonal matrix such that $det(A)<0$. Prove that there exists a nonzero vector $x\in\mathbb{R}^n$ such that $Ax=-x$.

Since $A$ is orthogonal and $det(A)<0$, its determinant is $-1$.

Please help, thanks a lot!

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    $\begingroup$ Nonzero vector, of course. Do you mean $Ax = -x$? Otherwise $-I$ in odd dimension is counterexample. $\endgroup$ – Robert Israel May 1 '18 at 18:50
  • $\begingroup$ @RobertIsrael Yes, I meant $Ax=-x$ and nonzero vector, Thank you! $\endgroup$ – user138017 May 1 '18 at 20:41
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$$\det(A+I)=\det(A)\det(A^{-1}+I)=-\det(A^\top+I)=-\det(A+I)$$

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