I've been going around in circles on this proof about Houseoholder reflectors. Proble
Let $x,y \in K^n$ such that $||x||_2=||y||_2$ and $x\neq y$:
a) If $K=\mathbb{R}$, prove that exists an unitary vector $\textbf{u}$ such that the corresponding Householder reflector be such that $R \textbf{x} = \textbf{y}$.
For this one I've proposed $\textbf{u}= \frac{\textbf{x-}\textbf{y}}{||\textbf{x-}\textbf{y}||_2}$. Then I literally computed:
$R\textbf{x}=I_n\textbf{x} -2 \textbf{uu*}\textbf{x}=I_n\textbf{x} -2\frac{\textbf{x-}\textbf{y}}{||\textbf{x-}\textbf{y}||_2} \frac{\textbf{x*-}\textbf{y*}}{||\textbf{x-}\textbf{y}||_2}\textbf{x}$ and after lot's of operations I arrived to nothing. Am I choosing incorrectly $\textbf{u}$?
b) If $K=\mathbb{C}$ prove that exists an unitary complex number $e^{i\theta}$ and an unitary vector $u$ such that the corresponding Householder reflector be such that $Rx=e^{i\theta}y$.
For this one, I ain't got a clue So any hint would be really appreciated. Thanks in advance.
You are on the right track. $u = \frac{x-y}{\lVert x-y\rVert_2}$ is the only sensible choice. Because of $\lVert x\rVert_2 = \lVert y\rVert_2$, we have $$ 0 = \lVert x\rVert_2^2 - \lVert y\rVert_2^2 \stackrel{(*)}= \langle x-y, x+y\rangle = \langle x-y, x\rangle + \langle x-y,y\rangle. $$ Hence $\langle x-y,x\rangle = \langle x-y,-y\rangle$ and therefore $$ 2\langle x-y,x\rangle = \langle x-y, x\rangle + \langle x-y, x\rangle = \langle x-y,x\rangle + \langle x-y,-y\rangle = \langle x-y,x-y\rangle. $$ Now, it follows easily that $$ x - 2\langle u,x\rangle u = x - \frac{2\langle x-y,x\rangle}{\langle x-y,x-y\rangle}\cdot (x-y) = x - (x-y) = y. $$
Edit: (b) follows in much the same way. Replace $y$ by $e^{{\rm i}\theta}y$ in the previous argumentation, where $e^{{\rm i}\theta} := \frac{\langle x,y\rangle}{\lvert\langle x,y\rangle\rvert}$ is such that ${\rm Im}\bigl(e^{-{\rm i}\theta}\langle x,y\rangle \bigr)=0$. In this case, $(*)$ still holds true.