What is the Householder matrix for complex vector space? For $\Bbb R^n$, Householder matrix $Q=I-2vv^T$ is an operator that maps a vector to its reflection across a hyperplane of normal $v$. 

The following is an illustration for Householder operator of a general inner product space.
  
For $\Bbb R^n$, use standard inner product, then we have $x-2<x,v>v=x-2vv^Tx=(I-2vv^T)x$, where $I-2vv^T$ is the Householder matrix. It satisfies the following
  

However, I am wondering what would be the Householder matrix for complex vector space $\Bbb C^n$? It looks like $I-2vv^H$ is not right ($H$ denotes conjugate transpose), because by my calculation it does not satisfy above problem 5.7.3. Wikipedia suggests 



I tried, and I might be wrong, but it does not seem to satisfy problem 5.7.3 either.

$Q(x+y)=(x + y) - \frac{{x - y}}{{{{\left\| {x - y} \right\|}^2}}}({(x - y)^H}(x + y) + \frac{{{x^H}(x - y)}}{{{{(x - y)}^H}x}}{(x - y)^H}(x + y))$
It looks impossible for $Q(x+y)=x+y$ unless $x^Hy=y^Hx$.

 A: Take a vector $w \in\mathbb C^n$. The Householder matrix $U_w$ is be defined as
$$U_w := I - \frac{2}{w^*w}ww^*$$
where $^*$ denotes hermitian adjoint.
However, complex householder matrices don't have the "reflector" property that real hoseholder matrices has, and is shown in Problem 5.7.3 in your question.
The closest we get, is making sure that it goes in one way. Theorem 2.1.13 from Horn & Johnson Matrix Analysis 2nd ed (2012) presents just this. They explicitly construct a matrix $U(y, x)$ with the property $U(y, x)x = y$, and that coincides with a real householder matrix when possible..

Theorem 2.1.13. Let $x, y ∈ \mathbb C^n$ be given and suppose that $||x|| = ||y|| > 0$. If $y = e^{iθ} x$ for some real $θ$, let $U(y, x) = e^{iθ} I_n$; otherwise, let $φ ∈ [0, 2π)$ be such that
$x^* y = e^{iφ}||x^*y||$ (take $φ = 0$ if $x^*y = 0$); let $w = e^{iφ}x − y$; and let $U(y, x) = e^{iφ}U_w$,
in which $U_w = I − \frac{2}{w^*w}ww^*$
is a Householder matrix.

Then $U(y, x)$ is unitary and essentially Hermitian, $U(y, x)x = y$, and $U(y, x)z ⊥ y $ whenever $z ⊥ x$. If $x$ and $y$ are real, then $U(y, x)$ is real orthogonal: $U(y, x) = I$ if $y = x$, and $U(y, x)$ is the
real Householder matrix $U_{x−y}$ otherwise.
