Jordan form of an orthogonal matrix

Question

Let $A$ be an orthogonal matrix.

Then there exists $Q$ also orthogonal such that $QAQ^* = D$ for some diagonal matrix $D$.

Following this post:Elements of SO(n) is block-diagonalizable

I'm not sure how we got that the Jordan from contains the block matrix $R_t $from $|\lambda|=1$ $$R_{t}= \left(\begin{array}{cc} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{array}\right).$$

Thanks in advance!

Answer 1

We start with a real square matrix $A$. If $A$ has an eigenvalue $a+bi$ where $a,b \in \mathbb R, b \ne 0$ then $a-bi$ is also an eigenvalue. Note tha your $b$ can be my $-b$. Suppose $\mathbf v$ is an eigenvector for the eigenvalue $a+bi$. Write $\mathbf v=\mathbf v^{\prime}+ \mathbf v^{\prime \prime}i$ where $\mathbf{v^{\prime},v^{\prime \prime}}$ are real. Let $P=[\mathbf{v^{\prime \prime}\vdots v^{\prime}}]$. Then $$AP=P\begin{bmatrix}a&-b\\b&a \end{bmatrix}$$ Note that if your $b$ is my $-b$,then $b$ and $-b$ will be interchanged.Thus $$P^{-1}AP=\begin{bmatrix}a&-b\\b&a \end{bmatrix}.$$ In polar form, $$\begin{bmatrix}a&-b\\b&a \end{bmatrix}=\begin{bmatrix}r\cos\theta&-r\sin\theta\\r\sin\theta&r\cos\theta \end{bmatrix}.$$ If $A$ is orthogonal, then every eigenvalue is on the unit circle, so $r=1.$

Answer 2

$$ \def\eqdef{\stackrel{\text{def}}{=}} $$ Because an orthogonal matrix $\ A\ $ is normal, has real entries, and eigenvalues of absolute value $\ 1\ $, it has a diagonalisation of the form \begin{align} U^\dagger&AU=\\ &\text{diag}\big(e^{it_1},e^{-it_1},e^{it_2},e^{-it_2},\dots,e^{it_r},e^{-it_r},1,1,\dots,1,-1,-1,\dots,-1\big)\ . \end{align} If $$ W\eqdef\frac{1}{\sqrt{2}}\pmatrix{1&-i\\-i&1} $$ it is easy to check that \begin{align} W^\dagger\pmatrix{e^{it}&0\\0&e^{-it}}W&=W^\dagger\pmatrix{\cos t+i\sin t&0\\0&\cos t-i\sin t}W\\ &=\pmatrix{\cos t&\sin t\\-\sin t&\cos t}\ . \end{align} Thus, if \begin{align} V&\eqdef\\ &\pmatrix{W&0_{2\times2}&\dots&0_{2\times2}&0_{2\times(n-2r)}\\ 0_{2\times2}&W &\dots&0_{2\times2}&0_{2\times(n-2r)}\\ \vdots&&\ddots&&\vdots\\ 0_{2\times2}& 0_{2\times2}& \dots& W &0_{2\times(n-2r)}\\ 0_{(n-2r) \times2}&0_{(n-2r) \times2}&\dots&0_{(n-2r) \times2}&I_{(n-2r) \times(n-2r)}}\ , \end{align} then conjugation of $\ U^\dagger AU\ $ by $\ V\ $: $$ V^\dagger U^\dagger AUV=(UV)^\dagger AUV $$ will turn each $\ 2\times2\ $ block $\ \pmatrix{e^{it_k}&0\\0&e^{-it_k}}\ $ on the diagonal of $\ U^\dagger AU\ $ into the corresponding $\ 2\times2\ $ block $\ \pmatrix{\cos t_k&\sin t_k\\-\sin t_k&\cos t_k}\ $ on the diagonal of $\ (UV)^\dagger AUV\ $.