# Finding every Householder reflection that converts $x$ to $y$, which are unit vectors.

## Problem

Find every Householder reflection $$H_v$$ (with respect to $$v$$), such that

$$y = H_vx$$

where $$x,y$$ are unit vectors, i.e. $$\Vert x \Vert_2 = \Vert y \Vert_2 = 1$$, and $$\langle x,y\rangle = \langle y,x\rangle$$ with $$x,y \in \mathbb{C}^n$$.

A Householder reflection with respect to $$v$$ is defined

$$H_v := I - 2\frac{vv^\ast}{v^\ast v}$$

which is the reflector with respect to the hyperplane orthogonal to $$v$$.

## Try

I have found one. By defining $$v := (x-y)$$, we have

$$H_v = I - 2 \frac{(x-y)(x-y)^\ast}{\Vert x-y \Vert_2^2}$$

and since $$\Vert x- y \Vert_2^2 = (x-y)^\ast (x-y) = 2 - 2x^\ast y$$ and $$(x-y)^\ast x = 1-x^\ast y$$, we have

$$H_v x = x - (x-y) = y$$

## Question

I would like to know if every $$H_v$$ is expressed

$$H_v = I - 2\frac{vv^\ast}{v^\ast v}$$

where $$v := c(x-y)$$ for some constant $$0 \neq c \in \mathbb{C}$$. In other words, I would like to verify the uniqueness of the above $$H_v$$.

But I cannot proceed from here. Any help will be appreciated.

## 1 Answer

If $$x\neq y$$, the Householder $$H$$ is unique. Any orthogonal (over $$\mathbb{R}$$) or unitary (over $$\mathbb{C}$$) reflections in a linear subspace is completely specified by its $$-1$$-eigenspace (or the $$+1$$-eigenspace, since they are orthogonal complements). For a Householder, you have the additional constraint $$\dim\ker(H+I)=1$$, and you know the $$x-y\neq 0$$ is in the $$-1$$-eigenspace.

If $$x=y$$, the choice of $$H$$ is in one-to-one correspondence with lines in $$x^\perp$$. In particular, there are infinitely many if $$n>2$$.

• Could you please translate the "$-1$-eigenspace" and "$+1$-eigenspace" into more plain notations..? – Moreblue May 21 at 22:53