Show that multiplication by a Householder matrix does not change the resulting vector's length. Suppose $H$ is an $n$ x $n$ Householder matrix.
Suppose that $w$ and $v$ are $n$-vectors with $w$ = $Hv$.
Assume $v = (v_1,v_2,...,v_n)^T$ and H is generated by a vector $u = (u_1, u_2, ..., u_n)^T$.
Show that $\left\lVert w \right\rVert = \left\lVert v \right\rVert$.
I've been given a hint, but I am not sure if it's required:
For $a = (a_1,a_2,...a_n)^T$ and $b = (b_1,b_2,...,b_n)^T$, $$a_1b_1 + a_2b_2 + ... + a_nb_n = a^Tb$$.
So far, I have calculated a few things that I know are true, but I am not sure how to proceed with them.


*

*$\left\lVert w \right\rVert =\sqrt {w_1^2 + w_2^2 + ... + w_n^2} = \sqrt{w^Tw}$

*$\left\lVert v \right\rVert = \sqrt {v_1^2 + v_2^2 + ... + v_n^2} = \sqrt{v^Tv}$

*$Hv = v - \frac{2u^Tv}{u^Tu}u$

*I also know that $H = H^T = H^{-1}$.


Can anyone get me rolling in the right direction with any of these?
 A: All we have to do is follow the calculation of $|| \vec w ||$.
$|| \vec w || = || H \vec v ||$ since $\vec w = H \vec v$. 
Next, we can see that $|| H \vec v || = \sqrt{ (H \vec v)^T (H \vec v) }$ by the definition of the length of a vector. 
$(H \vec v)^T = \vec v^T H^T$ since we can distribute the transpose operation, so $\sqrt{ (H \vec v)^T (H \vec v) } = \sqrt { \vec v^T H^T H \vec v }$. 
$H^TH = I$ by the properties of a Householder matrix, so $\sqrt { \vec v^T H^T H \vec v } = \sqrt { \vec v^T \vec v }$. 
Lastly, $\sqrt { \vec v^T \vec v } = || \vec v ||$ by the definition of the length of a vector, as we used before. 
Using all of this together, we can see that $|| \vec w || = || \vec v || $ is in fact true.
$$ || \vec w || = || H \vec v || = \sqrt{ (H \vec v)^T (H \vec v) } = \sqrt { \vec v^T H^T H \vec v } = \sqrt { \vec v^T \vec v } = || \vec v ||$$
A: We will show $\mathbf{H}$ is orthogonal matrix which implies Householder reflections preserve length.
Define the Householder matrix:
$$
\mathbf{H} = \mathbf{I} - 2\frac{u u^{T}}{u^{T} u}
$$
Show the Householder matrix is orthogonal:
$$
\begin{align}
\mathbf{H} \,\mathbf{H}^{T} &=
%
\left( \mathbf{I} - 2\frac{u u^{T}}{u^{T} u} \right)
%
\left( \mathbf{I} - 2\frac{u u^{T}}{u^{T} u} \right)^{T} \\
%
&=
%
\mathbf{I} - 4\frac{u u^{T}} {u^{T} u} + 4\frac{u \left(u^{T}u\right) u^{T}} {\left(u^{T} u\right)^{2}} \\
%
&= \mathbf{I} - 4\frac{u u^{T}} {u^{T} u} + 4\frac{u u^{T}} {u^{T} u} \\
%
&= \mathbf{I}
%
\end{align}
$$
As noted by @amd, the final step is to show Householder reflections preserve length. For a nonzero vector $x$,
$$
\begin{align}
  \lVert \mathbf{H}\, x \rVert^{2} &=
  \left( \mathbf{H}\, x \right)^{T} \left( \mathbf{H}\, x \right) \\
  &= x^{T} \left(\mathbf{H}^{T} \, \mathbf{H}\right) x \\
  &= x^{T} x.
\end{align}
$$ 
Therefore 
$$
\lVert \mathbf{H}\, x \rVert = \lVert x \rVert
$$
