Is orthogonal matrix diagonalizable?

Question

My understanding was that every normal matrix is diagonalizable. However I recently read that this applies to unitary but not necessary to orthogonal matrices.

"For every orthogonal operator of euclidean space there is an orthonormal basis in relation to which matrix of that operator is diag(+/-1,...,+/-1,A1,A2,...,Ak) where Ai are matrices of order 2."

Does this mean that orthogonal matrices aren't diagonalizable in $\Bbb R$?

Answer 1

Indeed, an orthogonal matrix doesn't have to be diagonalizable over $\Bbb R$. If it was, there would be a basis of eigenvators. No, consider, say, $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$. It has no eigenvector over $\Bbb R$. However, if we see it as the matrix of alinear map from $\Bbb C^2$ into itself, then $(1,i)$ is an eigenvector.

Answer 2

You might want to try diagonalizing $$\pmatrix{ c & -s \\ s & c} $$ where $c$ and $s$ are the sine and cosine of various angles. Trying it for $\pi/4$ should be enlightening.