isometries-orthogonal-complexify

Exercise.hardBeyond 21-242 Let $𝑄 \in O (𝑛)$ , and suppose $𝑄 𝐳 = 𝑒^{𝑖 𝜃} 𝐳$ for some $𝐳 = 𝐱 + 𝑖 𝐲 \in ℂ^{𝑛}$ with $𝐱, 𝐲 \in ℝ^{𝑛}$ and $𝜃 \notin 𝜋 ℤ$ .

(a)Show that $𝐳 ⟂ 𝐳$ , and deduce that $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ and $𝐱 ⟂ 𝐲$ .
(b)Show that $span (𝐱, 𝐲)$ is $𝑄$ -invariant, and find the matrix of $𝑄$ on it in the orthonormal basis $𝐱 / ‖ 𝐱 ‖, 𝐲 / ‖ 𝐲 ‖$ .
(c)Use (a) and (b) to give a third proof of Theorem 8.4.25.

Solution.

Uses. Theorem 8.4.15 (Spectral form of unitary operators), Theorem 8.1.18 (Eigenspaces of a normal operator are orthogonal), Proposition 7.3.34 (Eigenvalues of a complexification), Lemma 8.4.14 (Invariant complements)

Techniques. Complexify, then take real and imaginary parts (Technique 7.3.35)

Hint 1. Since

𝑄

is real,

𝑄 𝐳 = 𝑒^{- 𝑖 𝜃} 𝐳

. Eigenvectors of a normal matrix for distinct eigenvalues are orthogonal.

Hint 2.

𝐳 \cdot 𝐳 = \sum 𝑧_{𝑘}^{2}

. Expand with

𝑧_{𝑘} = 𝑥_{𝑘} + 𝑖 𝑦_{𝑘}

. For (b), compare real and imaginary parts of

𝑄 𝐳 = 𝑒^{𝑖 𝜃} 𝐳

Hint 3. For (c), induct: split off either a real eigenline or such a plane, and use Lemma 8.4.14.

Sketch. (a)

𝑄 𝐳 = 𝑒^{- 𝑖 𝜃} 𝐳

with

𝑒^{- 𝑖 𝜃} \neq 𝑒^{𝑖 𝜃}

, so

0 =

𝐳 \cdot

𝐳 =

\sum 𝑧_{𝑘}^{2} =

{‖ 𝐱 ‖}^{2} -

{‖ 𝐲 ‖}^{2} +

2 𝑖 𝐱 \cdot

𝐲

. (b)

𝑄 𝐱 =

cos 𝜃 𝐱 -

sin 𝜃 𝐲

𝑄 𝐲 =

sin 𝜃 𝐱 +

cos 𝜃 𝐲

; the matrix is

𝑅_{- 𝜃}

. (c) If

𝑄

has a real eigenvalue, split off its eigenline; otherwise split off the plane of (b). In both cases the complement is invariant, and induction applies.

(a) $𝑄$ has real entries, so conjugating $𝑄 𝐳 = 𝑒^{𝑖 𝜃} 𝐳$ gives $𝑄 𝐳 = 𝑒^{- 𝑖 𝜃} 𝐳$ , and $𝑒^{- 𝑖 𝜃} \neq 𝑒^{𝑖 𝜃}$ since $𝜃 \notin 𝜋 ℤ$ . As a complex matrix $𝑄$ is unitary, hence normal, so its eigenvectors for distinct eigenvalues are orthogonal (Theorem 8.1.18): $𝐳 \cdot 𝐳 = 0$ . Now

𝐳 \cdot 𝐳 = \sum_{𝑘} 𝑧_{𝑘} 𝑧_{𝑘} = \sum_{𝑘} 𝑧_{𝑘}^{2} = \sum_{𝑘} (𝑥_{𝑘}^{2} - 𝑦_{𝑘}^{2} + 2 𝑖 𝑥_{𝑘} 𝑦_{𝑘}) = {‖ 𝐱 ‖}^{2} - {‖ 𝐲 ‖}^{2} + 2 𝑖 (𝐱 \cdot 𝐲) .

Both the real and imaginary parts vanish: $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ and $𝐱 \cdot 𝐲 = 0$ . Since $𝐳 \neq 𝟎$ , both are nonzero.

(b) Write $𝑒^{𝑖 𝜃} = 𝑐 + 𝑖 𝑠$ with $𝑐 = cos 𝜃$ , $𝑠 = sin 𝜃$ . Then $𝑄 𝐱 +$ $𝑖 𝑄 𝐲 =$ $(𝑐 + 𝑖 𝑠) (𝐱 + 𝑖 𝐲) =$ $(𝑐 𝐱 - 𝑠 𝐲) +$ $𝑖 (𝑠 𝐱 + 𝑐 𝐲)$ . Comparing real and imaginary parts, $𝑄 𝐱 = 𝑐 𝐱 - 𝑠 𝐲$ and $𝑄 𝐲 = 𝑠 𝐱 + 𝑐 𝐲$ . So $span (𝐱, 𝐲)$ is invariant, and since $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ , the same formulas hold for the normalized vectors. The matrix is $(\begin{matrix} 𝑐 & 𝑠 \\ - 𝑠 & 𝑐 \end{matrix}) = 𝑅_{- 𝜃}$ , in agreement with Proposition 7.3.34(c).

(c) Induct on $𝑛$ , for an orthogonal operator $𝑆$ on a real inner product space $𝑉$ , using its matrix $𝑄$ in an orthonormal basis. If $𝑄$ has a real eigenvalue, it is $\pm 1$ , and a unit eigenvector spans an invariant line. Otherwise, by Theorem 8.4.15 $𝑄$ has an eigenvector $𝐳$ in $ℂ^{𝑛}$ for some $𝑒^{𝑖 𝜃}$ with $𝜃 \notin 𝜋 ℤ$ , and (b) gives an invariant plane on which $𝑆$ acts by $𝑅_{- 𝜃}$ in an orthonormal basis. In both cases Lemma 8.4.14 makes the orthogonal complement invariant, with orthogonal restriction, and induction gives the block form there. Normalizing the angles and ordering the blocks is as in the first proof of Theorem 8.4.25.