Find a method to use given matrices to create an orthogonal matrix $Q \in R^{n \times n}$, that for unitary vectors $x,y$ $\in R^{n}$, $$Q^Tx=y$$
The ideia that i have is: take a sucession of Givens matrices $G_n...G_1y=x$, the $G_n...G_1$ must be $Q$, how $Q$ is ortogonal matrix, just take the transpose to $Q^Tx=y$. But the problem is the Givens matrices.
It is better to use Householder transformations: The idea is $$ Q^T x = x - 2 (x^Tv)v. $$ with some $v$ with $\|v\|=1$. By the condition $$ y = Q^Tx = x-2 (x^Tv)v $$ it follows that $v$ is a multiple of $x-y$. Set $v:= \frac1{\|x-y\|}(x-y)$. It remains to show that the matrix defined by $$ Q^Tx = x-2 (x^Tv)v $$ indeed does the job: $$ \begin{split} Q^Tx-y &= x-y - \frac2{\|x-y\|^2} x^T(x-y)\cdot (x-y)\\ & = (x-y) \|x-y\|^2\left( \|x-y\|^2 - 2x^T(x-y)\right)\\ & = (x-y) \|x-y\|^2\left( -(x+y)^T(x-y)\|x-y\|^2 \right)\\ & = (x-y) \|x-y\|^2\left(\|y\|^2 - \|x\|^2 \right) = 0. \end{split}$$