How to find a suitable orthogonal matrix Assume $x,y \in \text{R}^n$ are two vectors of the same length, how to prove that there is an orthogonal matrix $A$ such that $Ax=y$?
Thanks for your help.
 A: You can explicitly use a Householder matrix. Normalise $x$ and $y$ so that they become unit vectors (the special case $x=y$ is trivial and is ignored here). Let $u = x-y$ and $Q = I - \frac{uu^\top}{x^\top u}$. Then $Q$ is a real symmetric and orthogonal matrix such that $Qx=y$.
A: Well, if they both have length zero, then take $A$ to be anything. If $\|x\| = \lambda = \|y\|$, then divide by $\lambda$ and assume that $\lambda = 1$.
Now note that the relation $x \sim y$ iff $\exists$ an orthogonal matrix $A$ such that $Ax = y$ is an equivalence relation. So it is enough to assume that $x = (1,0,0,\ldots, 0)$.
Now take $y$ and extend it to an orthonormal basis $\mathcal{B}$ of $\mathbb{R}^n$. Consider the change of basis matrix $A$ that sends the standard basis to $\mathcal{B}$. The columns of this matrix are precisely the vectors of $\mathcal{B}$ which are orthonormal, and hence $A$ is an orthogonal matrix. Furthermore,
$$
Ax = y
$$
A: Since every orthogonal matrix $A$ always has inverse, $$ AA^{t} = A^{t}A = I $$, so that if $A$ is orthogonal then any equation $Ax=y$ has a solution $$ x = (...)y $$
