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If $P$ is real orthogonal matrix with $\det P = -1$, prove that $-1$ is an eigenvalue of $P$.

can anyone help me please how can I solve this problem?

Michael Hardy
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user67634
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3 Answers3

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Hint:

What are the possible eigenvalues of $P$ ?

If $-1$ is not an eigenvalue, what must be the eigenvalues of $P$ ? hence $\det(P)=?$

Michael Hardy
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Belgi
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  • Since the entries in the matrix are real, the non-real eigenvalues come in complex-conjugate pairs, and since the matrix is orthogonal, the eigenvalues have norm $1$. And every complex number of norm $1$ occurs as an eigenvalue of some orthogonal matrix. If $-1$ is not an eigenvalue, then the only real eigenvalues are $1$ and the determinate is $1$. However, I don't think everything above is so obvious just from the definition that I'd leave it as an exercise like this. – Michael Hardy Mar 20 '13 at 11:58
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Belgi has already hinted at the standard way of solving this problem. More specifically, suppose $\lambda_1,\ldots,\lambda_n\in\mathbb{C}$ are the eigenvalues of $P$.

  1. Every real eigenvalue of $P$, if exists, must be equal to $-1$ or $1$.
  2. As $P$ is real, nonreal eigenvalues of $P$, if exist, must occur in conjugate pairs, the product of all nonreal eigenvalues of $P$ is a positive real number.
  3. Hence $-1$ is an eigenvalue of $P$ because $\det P$, the product of all eigenvalues of $P$, is negative.

However, there are alternative proofs as well. Here is one that makes use of Cayley transform. Suppose $-1$ is not an eigenvalue of $P$. By Cayley transform, $A=(I-P)(I+P)^{-1}$ is a well defined skew symmetric matrix such that $P=(I-A)(I+A)^{-1}$. Hence $$ \det P=\det\left[(I-A)(I+A)^{-1}\right]=\det(I-A)^T\det(I+A)^{-1} =\det(I+A)\det(I+A)^{-1}=1, $$ which contradicts the condition that $\det P=-1$. Therefore $-1$ must be an eigenvalue of $P$.

user1551
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An orthogonal matrix is diagonalizable (in an orthonormal basis, moreover) and its eigenvalues are in $\{\pm 1\}$. You see what happens if the determinant, which is the product of the eigenvalues with multiplicities, is $-1$. This means that $-1$ has odd multiplicity, in particular $\geq 1$.

Julien
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