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I want to show that if a real orthogonal matrix $A$ has determinant $-1$ then $\lambda=-1$ must be an eigenvalue of $A$.

I have proven this in a long-winded way and I was wondering if these is a quick way of seeing it.

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The matrix is diagonalizable over $\mathbb C$, so the determinant is the product of the eigenvalues.

The complex eigenvalues come in conjugate pairs, and the product of two conjugate eigenvalues is a positive real. So there has to be at least one negative real eigenvalue.

The only negative real that can be an eigenvalue of an orthogonal matrix (which preserves the Euclidean norm of a vector) is $-1$.


Actually it's not necessary to appeal to diagonalizability; just considering the characteristic polynomial will do.

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  • $\begingroup$ Ahhh, I forgot about the conjugate pairs. Thanks. $\endgroup$ – user85798 May 12 '18 at 22:14
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If $A$ has $-1$ has a determinant, it is a symmetry, since a rotation preserves the orientation. Then $\mathbb{R}^n=U\oplus V$ where $U=\{x:A(x)=x\}$, $V=\{x:A(x)=-x\}$.

You also have $AA^T=I$, $det(A+I)=det(A(I+A^{-1}))=det(A)det(I+A^T)=-det(I+A^T)$.

$(I+A)^T=(I+A^T)$ implies that $det(I+A)=det(I+A^T)$, you deduce that $det(I+A)=-det(I+A)$ and $det(I+A)=0$.

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