30 - Canonical Form of Orthogonal Operators

Canonical Form of Orthogonal Operators

Let $T:V\to V$ be orthogonal and $V$ be a finite dimensional real vector space then

There exists orthogonal basis $\mathcal{B}$ such that

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\left[\begin{array}{ccccc}I& & & & \\ & -I& & & \\ & & R_{{\theta }_1}& & \\ & & & \ddots & \\ & & & & R_{{\theta }_l}\end{array}\right]\text{ where }{\theta }_i\ne 0,\pi$$

where

$$R_{\theta }=\left(\begin{array}{cc}\cos (\theta )& -\sin (\theta )\\ \sin (\theta )& \cos (\theta )\end{array}\right)$$

Proof

Let $S=T+T{1}^{-1}=T+T^{\ast }$ so $S^{\ast }=(T+T^{\ast }{)}^{\ast }=T^{\ast }+T=S$
Hence $S$ is self-adjoint so it has a Orthonormal Basis for Self-Adjoint Maps thus

$$V=V_1\oplus \cdots \oplus V_k$$

decomposes into orthogonal eigenspaces of $S$ with distinct eigen values ${\lambda }_1,\cdots ,{\lambda }_k$
As each $V_i$ is $T-$invariant for $v\in V_i$ so

$$S(T(V))=T(S(v))={\lambda }_{i}T(v)\text{ and so }T(v)\in V_i$$

So restrict to $T{\mid }_{V_i}$

By definition of $V_i$ for all $v\in V_i$ then

$$(T+T^{-1})v={\lambda }_{i}v$$

Hence $T^2-{\lambda }_{i}T+I=0$ hence the minimal polynomial of $T{\mid }_{V_i}$ divides $x^2-{\lambda }_{i}x+1$
So any eigenvalue of $T{\mid }_{V_i}$ is a root of it

If ${\lambda }_i=\pm 2$ then $(T+I{)}^2=0$ or $(T-I{)}^2=0$
Thus the eigenvalue of $T{\mid }_{V_i}$ is $-1$ or $+1$ respectively
Since we know $T{\mid }_{V_i}$ may be diagonalised over $\mathbb{C}$ then

$$T{\mid }_{V_i}=+I\text{ or }-I$$

If ${\lambda }_i\ne \pm 2$ then $T{\mid }_{V_i}$ does not have any real eigenvalues as they would to be $\pm 1$ by eigenvalues of unitary transformations
Hence this forces ${\lambda }_i=\pm 2$

So$\{v,T(v)\}$ are linearly independent over the reals for $v\ne 0\in V_i$

Consider plane $W=⟨v,T(v)⟩$ 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 $T$-invariant by Orthogonal Complement of Invariant Subspaces for Unitary Operators
So we see that $V_i$ splits into $2$-dimensional $T-$invariant subspaces hence

$$T{\mid }_W=R_{\theta }$$

for some orthonormal basis of $W$ and for some $\theta \ne 0,\pi$