An orthogonal matrix in $\mathbb{R}^{3\times3}$ with real eigenvalues is diagonalizable

Question

I know there are two non trivial (i.e. if we solve these two cases the other cases are trivial) cases:

$\lambda_{1,2,3}=1$

and:

$\lambda_1=1,\lambda_{1,2}=-1$

I have been trying to use generalized eigenvectors and the Jordan Canonical and the fact that the JCF of matrix A can be rewritten as:

$\hat{A}=\hat{D}+\hat{N}$

Then I assume that $\hat{N}$ is not the zero matrix, but I don't know how/where to invoke the fact that A is orthogonal to get a contradiction. It feels like I have some of the pieces but not all of them...Any help would be greatly appreciated. This question is specific orthogonal matrices with the aforementioned eigenvalues. It would be helpful if this question was solved using definitoins for generalized eigenvectors and the JCF.

Answer 1

Alright I think I figured it out. If A has a nontrivial JCF then we have an egeinvalue $\lambda\in\{-1,1\}$ and two nonzero vectors $v,w\in\mathbb{R}^3$ such that

$Av=\lambda v+w$

$Aw=\lambda w$

These come from the first two columns of any nontrivial Jordan Block, and also get rid of the cases I was talking about earlier. They are also equivalent to assuming that $A$ is not diagonalizable. Dot the first equation with $w$, and use the fact that $A$ is orthogonal:

$\lambda(v,w)+(w,w)=(Av,w)=(v,A^Tw)=(v,A^{-1}w)$

From the second equation we have that:

$Aw=\lambda w $\iff$ A^{-1}w=\lambda^{-1}w=\lambda w$

Therefore we get:

$\lambda(v,w)+(w,w)=\lambda(v,w)$

which implies that

(w,w)=0

thus we have a contradiction, and A must be diagonalizable

Answer 2

Let $A$ be your matrix, let $e_1$ be an eigenvector of $A$, and let $V=e_1^\perp$. Then, since $A$ is orthogonal, $A.V\subset V$. But $V$ is $2$-dimensional and the restriction of $A$ to $V$, seen as a linear map from $V$ to $V$, also only has real eigenvalues and it is also orthogonal. Therefore, it is diagonalizable: you ake an eigenvector $e_2$ of $A|_V$ and then, if $e_3$ is a non-nul vector of $V$ which is orthogonal to $e_2$, then $A.\operatorname{span}\{e_3\}\subset\operatorname{span}\{e_3\}$; in other words, $e_3$ is also an eigenvector of $A$. And then $\{e_1,e_2,e_3\}$ is a basis of eigenvectors of $A$.