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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.

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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 $n>2$.

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

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