Given an $n$-vector $x$, show that floating-point computation of the Householder vector $v$ such that $P x = (I − 2vv^{T} )x = \pm\left\|x\right\|_{2}e_{1} $ gives a forward stable result $v^{\prime}$ satisfying $\left\|v^{\prime} − v\right\|_{2} \le \mathcal O(\varepsilon)\left\|v\right\|_{2}$. Is it also backward stable? Prove or give a counterexample
I have no idea how to begin this work. Thanks in advance!
