the number of distinct real spectra for an orthogonal $n×n$ matrix

hbghlyj

$n=2$	$\pmatrix{1&0\\0&1},\pmatrix{1&0\\0&-1},\pmatrix{-1&0\\0&-1},\pmatrix{\cosθ&-\sinθ\\\sinθ&\cosθ}$
$n=3$	$\pmatrix{1&0&0\\0&1&0\\0&0&1},\pmatrix{1&0&0\\0&1&0\\0&0&-1},\pmatrix{1&0&0\\0&-1&0\\0&0&-1},\pmatrix{-1&0&0\\0&-1&0\\0&0&-1},$$\pmatrix{1&0&0\\0&\cosθ&-\sinθ\\0&\sinθ&\cosθ},\pmatrix{-1&0&0\\0&\cosθ&-\sinθ\\0&\sinθ&\cosθ}$

A002620:  Quarter-squares: $a_n = \lfloor n/2\rfloor·\lceil n/2\rceil$. Equivalently, $a_n =\lfloor n^2/4\rfloor$.

a(n+2) is the number of distinct real spectra (eigenvalues repeated according to their multiplicity) for an orthogonal n X n matrix. The case of an empty spectrum list is logically counted as one of those possibilities, when it exists. Thus a(n+2) is the number of distinct reduced forms (on the real field, in orthonormal basis) for elements in O(n). - Christian Devanz, Feb 13 2019

decomposition of orthogonal operators as rotations and reflections
Orthogonal matrix#Canonical form

More broadly, the effect of any orthogonal matrix separates into independent actions on orthogonal two-dimensional subspaces. That is, if $Q$ is special orthogonal then one can always find an orthogonal matrix $P$, a (rotational) change of basis, that brings $Q$ into block diagonal form:
$$P^\mathrm{T}QP = \begin{bmatrix}
R_1 & & \\ & \ddots & \\ & & R_k
\end{bmatrix}\ (n\text{ even}),
\ P^\mathrm{T}QP = \begin{bmatrix}
R_1 & & & \\ & \ddots & & \\ & & R_k & \\ & & & 1
\end{bmatrix}\ (n\text{ odd}).$$where the matrices $R_1,\dots,R_k$ are $2 × 2$ rotation matrices, and with the remaining entries zero. Exceptionally, a rotation block may be diagonal, $±I$. Thus, negating one column if necessary, and noting that a $2 × 2$ reflection diagonalizes to a $+1$ and $−1$, any orthogonal matrix can be brought to the form
$$P^\mathrm{T}QP = \begin{bmatrix}
\begin{matrix}R_1 & & \\ & \ddots & \\ & & R_k\end{matrix} & 0 \\
0 & \begin{matrix}\pm 1 & & \\ & \ddots & \\ & & \pm 1\end{matrix} \\
\end{bmatrix},$$The matrices $R_1,⋯,R_k$ give conjugate pairs of eigenvalues lying on the unit circle in the complex plane; so this decomposition confirms that all eigenvalues have absolute value $1$. If $n$ is odd, there is at least one real eigenvalue, $+1$ or $−1$; for a $3 × 3$ rotation, the eigenvector associated with $+1$ is the rotation axis.

hbghlyj

A0.pdf

Theorem 8.32. Let $T : V → V$ be orthogonal and $V$ be a finite dimensional real vector space. Then there exists an orthonormal basis $\cal B$ such that:$$\sideset{_\mathcal{B}}{_\mathcal{B}}{[T]}=\left[\begin{array}{ccccc}I & & & & \\ & -I & & & \\ & & R_{\theta_{1}} & & \\ & & & \ddots & \\ & & & & R_{\theta_{\ell}}\end{array}\right] \qquad \theta_{i} \neq 0, \pi$$Proof. Let $S=T+T^{-1}=T+T^{*}$. Then $S^{*}=\left(T+T^{*}\right)^{*}=T^{*}+T=S$. So $S$ is self-adjoint and has a basis of orthonormal eigenvectors by Theorem 8.21 and thus $$ V=V_{1} \oplus \cdots \oplus V_{k} $$ decomposes into orthogonal eigenspaces of $S$ with distinct eigenvalues $\lambda_{1}, \ldots, \lambda_{k}$. Note that each $V_{i}$ is also $T$-invariant as for $v \in V_{i}$ $$ S(T(v))=T(S(v))=\lambda_{i} T(v) \text { and so } T(v) \in V_{i} $$ So we may restrict ourselves to $\left.T\right|_{V_{i}}$.

By definition of $V_{i}$, for all $v \in V_{i}$ we have $\left(T+T^{-1}\right) v=\lambda_{i} v$ and hence $T^{2}-\lambda_{i} T+I=0$. Thus the minimal polynomial of $\left.T\right|_{V_{i}}$ divides $x^{2}-\lambda_{i} x+1$ and any eigenvalue of $\left.T\right|_{V_{i}}$ is a root of it.

If $\lambda_{i}=\pm 2$, then $\left(T+\lambda_{i} I\right)^{2}=0$ or $\left(T-\lambda_{i} I\right)^{2}=0$. Thus the only eigenvalue of $\left.T\right|_{V_{i}}$ is $-1$ or $+1$, respectively. Since we know $\left.T\right|_{V_{i}}$ may be diagonalised over $\mathbb{C}$ (Theorem 8.29), we must have $\left.T\right|_{V_{i}}=+I$ or $-I$.

If $\lambda_{i} \neq \pm 2$ then $\left.T\right|_{V_{i}}$ does not have any real eigenvalues as they would have to be $\pm 1$ by Lemma 8.26 (with product $+1$ ) forcing $\lambda_{i}=\pm 2$. So $\{v, T(v)\}$ are linearly independent over the reals for $v \neq 0 \in V_{i}$. Consider the plane $W=\langle v, T(v)\rangle$ spanned by $v$ and $T v$. Then $W$ is $T$-invariant as $$ v \mapsto T(v), T(v) \mapsto T^{2}(v)=\lambda_{i} T(v)-v . $$ Hence $W^{\perp}$ is also $T$-invariant by Lemma 8.28. Repeating the argument for $\left.T\right|_{W^{\perp}}$ if necessary, we see that $V_{i}$ splits into 2-dimensional $T$-invariant subspaces. By our Example 8.31, with respect to some orthonormal basis of $W$ $$ \left.T\right|_{W}=R_{\theta}=\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{array}\right] $$ for some $\theta \neq 0, \pi$. (Note, the fact that $\left.T\right|_{W}$ does not have any real eigenvalues implies that $\left.T\right|_{W}$ is not a reflection and $\theta \neq 0, \pi$.)

Footnote

One can also deduce this theorem just from our spectral theorem for unitary matrices (Theorem 8.29), by grouping the non-real eigenvalues in complex conjugate pairs $λ$ and $λ$ and taking an orthonormal basis of “real vectors” for each of the two dimensional spaces you get by choosing an eigenvector for $λ$ and the conjugate one for $λ$.