I know that the eigenvalues of the Householder reflector in $\mathbb{R^{m\times n}}$, $H=I-2qq^T$, is $\pm 1$.
But I have no idea whether this statement is true for the Householder reflector in $\mathbb{C^{m\times n}}$, $F=I-2qq^*$ where $*$ is Hermitian transpose.
Here is my short observation:
The eigenvalue $\lambda$ satisfies $Fx=\lambda x \Rightarrow x-2qq^*x=\lambda x$.
If $x \in \perp q:=\{z \vert q^*z=0\}$, then $x=\lambda x \Rightarrow \lambda=1$.
On the other hand, if $x=\mu q$ for some constant $\mu \in \mathbb{C}$, then $\mu q-2\mu q=\lambda \mu q$.
Then $\lambda = -1$.
I am curious at this point: Can we say that $\pm 1$ are the only eigenvalues of $F$?
If we are in $\mathbb{R}$, it is true since $F^2=I-4qq^T+4qq^T qq^T=I \Rightarrow \det(F)=\pm 1$.
But I think it is not that clear if we talk about complex $F$.
Could you give me more explanation?
Here are more observations:
Since $F$ is Hermitian, all eigenvalues of $F$ are real.
pf) If $Ax=\lambda x$ for some $x \in \mathbb{C} -\{0\}$, then $\lambda \Vert x \Vert^2=\lambda x^*x=x^*(\lambda x)=x^*Ax=x^*A^*x=(Ax)^*x=(\lambda x)^*x=\lambda^*x^*x=\lambda^*\Vert x \Vert^2$. Since $x \neq 0$, it must be the case of $\lambda=\lambda^*$, i.e., $\lambda \in \mathbb{R}$.
Since $F^2=I$, $\det(F)=\pm 1$.
pf) $\det(F^2)=\det(I) \Rightarrow (\det(F))^2=1 \Rightarrow \det(F)=\pm 1$.
Now my question is more clear: Can we assure that the eigenvalues of $F$ are $\pm 1$ if $\det(F)=\pm 1$ is given?
The fact that $F^2 = I$ tells us that all eigenvalues of $F$ (real or complex) are equal to $1$ or $-1$. Indeed, suppose that $\lambda \in \Bbb C$ is an eigenvalue. Let $x \in \Bbb C^n$ be an associated eigenvector (so that $x \neq 0$ and $Fx = \lambda x$). We have $$ x = Ix = F^2x = F(Fx) = F(\lambda x) = \lambda F(x) = \lambda^2x, $$ so that $\lambda^2 x = x$. Because $x \neq 0$, this implies that $\lambda^2 = 1$. It follows that $\lambda = 1$ or $\lambda = -1$.