# Thread: Eigenvalues of Orthogonal Matrices

1. ## Eigenvalues of Orthogonal Matrices

Show that if M is orthogonal, then M admits at most one eigenvalue, and as such must be either +1 OR -1.

Using the definition of orthogonality and eigenvalues, it's easy enough to show that the eigenvalues of M are +/- 1. I'm however unable to show that M admits at most one eigenvalue.

2. Originally Posted by h2osprey
Show that if M is orthogonal, then M admits at most one eigenvalue, and as such must be either +1 OR -1.

Using the definition of orthogonality and eigenvalues, it's easy enough to show that the eigenvalues of M are +/- 1. I'm however unable to show that M admits at most one eigenvalue.

This is false: $M=\left(\begin{array}{rrr}1&0&1\\\!\!0&\!\!-1&\!\!0\\0&0&1\end{array}\right)$ is orthogonal and its characteristic polynomial is $p_M(x)=(x-1)^2(x+1)\,\Longrightarrow$ both 1, -1 are eigenvalues of M.

Tonio

3. No, Tonio. M is not orthogonal. Two equivalent definitions of "orthogonal matrix" are:
1) It's columns are orthogonal vectors.
That is not true for your matrix, M, because the dot products of the first and third columns is 1, not 0.

2) The transpose equals the inverse matrix.
That is not true for your matrix, M, because
$\begin{pmatrix}1 & 0 & 0 \\0 & -1 & 0 \\ 1 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2\end{pmatrix}$.

4. I guess the 1 in the top right corner is a 0, just a little misprint, no?

5. Originally Posted by clic-clac
I guess the 1 in the top right corner is a 0, just a little misprint, no?

Indeed...good sight! Thanx

Tonio

6. Originally Posted by HallsofIvy
No, Tonio. M is not orthogonal. Two equivalent definitions of "orthogonal matrix" are:
1) It's columns are orthogonal vectors.
That is not true for your matrix, M, because the dot products of the first and third columns is 1, not 0.

2) The transpose equals the inverse matrix.
That is not true for your matrix, M, because
$\begin{pmatrix}1 & 0 & 0 \\0 & -1 & 0 \\ 1 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2\end{pmatrix}$.

Typo: the 1-3 entry should be 0. Thanx

Tonio

7. Originally Posted by h2osprey
Show that if M is orthogonal, then M admits at most one eigenvalue, and as such must be either +1 OR -1.

Using the definition of orthogonality and eigenvalues, it's easy enough to show that the eigenvalues of M are +/- 1. I'm however unable to show that M admits at most one eigenvalue.
What about $M=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$?

What about the matrix of a rotation of angle $\theta$? It has eigenvalues $e^{i\theta},e^{-i\theta}$. But perhaps you were dealing with real eigenvalues?

8. Thanks for all the help! It didn't ever occur to me that the question could be wrong, but obviously the counterexample was simple enough.