isometries-unit-vectors-transitive

Exercise.standard Let $𝐱, 𝐲 \in ℂ^{𝑛}$ with $‖ 𝐱 ‖ = ‖ 𝐲 ‖ \neq 0$ .

(a)Show that there is $𝑐 \in ℂ$ with $| 𝑐 | = 1$ such that $𝐱 \cdot (𝑐 𝐲)$ is real.
(b)Deduce that $𝐲 = 𝑐 𝐻 𝐱$ for some matrix $𝐻$ that is either $𝐼_{𝑛}$ or a Householder reflection. In particular a unitary matrix maps $𝐱$ to $𝐲$ .
(c)Find such a unitary for $𝐱 = (1, 0)$ and $𝐲 = (0, 𝑖)$ .

Uses. Proposition 8.4.21 (A reflection exchanging two vectors), Exercise 6.3.37, Proposition 6.1.4 (Sesquilinearity)

Hint 1.

𝐱 \cdot

(𝑐 𝐲) =

𝑐 (𝐱 \cdot 𝐲)

. Choose the phase of

𝑐

to cancel that of

𝐱 \cdot 𝐲

Hint 2. If

𝑐 𝐲 = 𝐱

take

𝐻 = 𝐼_{𝑛}

; otherwise apply Proposition 8.4.21 to

𝐱

and

𝑐 𝐲

Sketch. (a) If

𝐱 \cdot 𝐲 = 𝑟 𝑒^{𝑖 𝜑}

, take

𝑐 = 𝑒^{𝑖 𝜑}

𝐱 \cdot

(𝑐 𝐲) =

𝑐 𝑟 𝑒^{𝑖 𝜑} =

𝑟

. (b)

𝑐 𝐲

has the same length as

𝐱

and a real dot product with it, so

𝐻 𝐱 = 𝑐 𝐲

for

𝐻 = 𝐼_{𝑛}

or a Householder

𝐻

; then

𝑐 𝐻

is unitary. (c)

𝐱 \cdot 𝐲 = 0

, so

𝑐 = 1

works;

𝐝 = (1, - 𝑖)

and

𝐻 =

𝐼_{2} -

𝐝 𝐝^{*} =

(\begin{matrix} 0 & - 𝑖 \\ 𝑖 & 0 \end{matrix})

maps

(1, 0)

(0, 𝑖)

(a) By Proposition 6.1.4, $𝐱 \cdot$ $(𝑐 𝐲) =$ $𝑐 (𝐱 \cdot 𝐲)$ . If $𝐱 \cdot 𝐲 = 0$ , any $𝑐$ works. Otherwise write $𝐱 \cdot 𝐲 = 𝑟 𝑒^{𝑖 𝜑}$ with $𝑟 > 0$ , and let $𝑐 = 𝑒^{𝑖 𝜑}$ ; then $𝑐 𝑟 𝑒^{𝑖 𝜑} = 𝑟$ is real.

(b) Let $𝐲^{'} = 𝑐 𝐲$ , so $‖ 𝐲^{'} ‖ = ‖ 𝐱 ‖$ and $𝐱 \cdot 𝐲^{'} \in ℝ$ . If $𝐲^{'} = 𝐱$ , let $𝐻 = 𝐼_{𝑛}$ . Otherwise Proposition 8.4.21 gives a Householder reflection $𝐻$ with $𝐻 𝐱 = 𝐲^{'}$ . In both cases $𝑐 𝐻 𝐱 = 𝑐 𝑐 𝐲 = 𝐲$ , and $𝑐 𝐻$ is unitary, as a product of the unitary matrices $𝑐 𝐼_{𝑛}$ and $𝐻$ (Exercise 6.3.37).

(c) Here $𝐱 \cdot 𝐲 = 1 \cdot 0 + 0 \cdot 𝑖 = 0$ , so $𝑐 = 1$ works and $𝐲^{'} = 𝐲 = (0, 𝑖)$ . With $𝐝 = 𝐱 - 𝐲 = (1, - 𝑖)$ , ${‖ 𝐝 ‖}^{2} = 2$ , and

𝐻 = 𝐼_{2} - 𝐝 𝐝^{*} = 𝐼_{2} - (\begin{matrix} 1 & 𝑖 \\ - 𝑖 & 1 \end{matrix}) = (\begin{matrix} 0 & - 𝑖 \\ 𝑖 & 0 \end{matrix}) .

Then $𝐻 (1, 0) = (0, 𝑖) = 𝐲$ , as required, and $𝐻$ is a unitary matrix mapping $𝐱$ to $𝐲$ .