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I was asked the following question on a test:

If $O\in M_{3}(\mathbb{R})$ is orthogonal and $\det O=-1$ then $\lambda=-1$ is an eigenvalue of $O=(o_{ij})$ .

I tried building equations using $OO^{t}=I\ \Rightarrow[OO^{t}]_{ij}=\sum_{k=1}^{3}o_{ik}o_{jk}=\delta_{ij} $ and the statement about the determinant, but without success.

Can anyone give me a hint?

Thanks!

Paz
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    The determinant is the product of the eigenvalues. What do you know about the eigenvalues of orthogonal matrices? – celtschk Jul 14 '13 at 10:10
  • Is that the general case for orthogonal matrices with real values? I thought $O$ must be symmetric, thus ortho-diagnolizable, for that to happen. In that case $\lambda=\pm1$. – Paz Jul 14 '13 at 10:17
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    The eigenvalues of a general orthogonal matrix may be complex. Do you already know the properties of unitary matrices? Then you can obtain the properties of real orthogonal matrices by noting that they are just unitary matrices with all entries real. – celtschk Jul 14 '13 at 10:36
  • I understood what you meant. Thanks. – Paz Jul 14 '13 at 11:24

2 Answers2

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The only fact we need about orthogonal matrices is that their eigenvalues have absolute value 1.

Suppose all the eigenvalues are real. Then each eigenvalue has to be $1$ or $-1$. Since their product is $-1$, at least one needs to be $-1$.

If not all eigenvalues are real, there exists one complex eigenvalue $\lambda$. But then also $\overline{\lambda}$ is an eigenvalue. Let $\mu$ be the remaining eigenvalue. It follows that $$-1=\det O=\lambda \overline{\lambda} \mu=|\lambda|^2\mu=\mu$$

Anonymous999
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  • "But then also $\bar{\lambda}$ is an eigenvalue" because the original matrix has real entries? – Paz Jul 14 '13 at 12:05
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There is nothing wrong with Anonymous999's +1 answer. Just adding another approach, possibly a bit more holistic :-)

We have $$ \det (O+I)=\det(O+OO^t)=\det (O(I+O^t))=\det O \cdot \det(I+O^t)=-\det(I+O), $$ because taking the transpose doesn't change the determinant and it was given that $\det O=-1$.

As we are working over reals, this implies that $\det(I+O)=0$, and thus also $\det(-I-O)=0$. Therefore $\lambda=-1$ is a zero of the characteristic equation $\det(xI-O)=0$, and we are done.

Jyrki Lahtonen
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  • Nice, not sure I would have thought of that optimal approach since it is so tempting to do otherwise, +1. – Julien Jul 14 '13 at 13:49
  • I like this answer, but also wouldn't think of doing that. – Paz Jul 14 '13 at 17:47
  • This is an easier variant of the oft seen proof that $O$ has eigenvalue $1$, if $\det O=1$ and $n$ is odd. In that case we need an extra step telling $\det(O-I)=-\det(I-O)$. – Jyrki Lahtonen Jul 14 '13 at 21:07