Jordan form of an orthogonal matrix

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!

2 Answers

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 $$r=1.$$

$$\def\eqdef{\stackrel{\text{def}}{=}}$$ 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\$$.