How can we recover $u$ from a Householder matrix?

Question

Say, we have a Householder matrix $H$ (i.e. $H=H^T=H^{-1}$) and would like to find $u$ such that $$H=I-\frac{2}{u^Tu}uu^T.$$ Geometrically, it seems like a good idea to take some $x$ and try $u:=H(x)-x$ (as long as it's not zero). So I plugged in and arrived at $$I+\frac{1}{x^THx}(Hxx^TH-Hxx^T-xx^TH+xx^T),$$ which I don't know how to simplify further. If the above makes sense, how can we arrive at $H$?

Answer 1

Note that $\frac{1}{2}(I-H)$ is an orthogonal projection into a onedimensional subspace. Hence, take an arbitrary vector $x$ such that $\frac{1}{2}(I-H)x\ne 0$. Then $u=\frac{1}{2}(I-H)x$.

Answer 2

We have $Hu=u-2u=-u$. Since $H$ is known, we can find $u$ as the eigenvector corresponding to the eigen value $-1$ of $H$.