Question involving Householder Matrix This is an exam question that I'm having trouble solving.
Given a unit vector $v^Tv=1$, the Householder matrix is defined as $H=I-2vv^T$.
The first question is: given column vector $x$, if $Hx=c\cdot e_1$ where $c$ is constant and $e_1$ is the first vector of the canonical basis, find $c$.
After some algebra, I found that
$$c=x_1-2v_1\sum_{i=1}^n{v_ix_i}$$
I was unable to simplify it further.
The second question that I've been unable to solve is:
What is $v$, so that $H=I-2vv^T$ satisfies $Hx=c\cdot e_1$?

It might be a little simpler to look at what $H$ does. If $x || v$ then
$Hx = -x$ and if $x \bot v$, we have $Hx = x$. So to invert, we just apply $H$ again.
To confirm, check that $H^2 = I$, in fact, $H$ is orthogonal, which gives it
desirable numerical properties.
I suspect the purpose of the question was not to have you perform the 
algebraic manoeuvre $c=e_1^T Hx= e_1^T(x-2v^Tx v) = x_1 - 2v_1(v^T x)$, but to note that
$\|Hx\| = \|ce_1\|$ from which we get $c = \pm \|x\|$.
To compute a relevant $v$, note that we want to reflect the vector $x$
onto the vector $\pm \|x\| e_1$. Use $y=\pm e_1$ to represent the
desired target unit vector. Then we want to find the hyperplane that
 bisects the angle between $x$
and $\|x\| y$,
we can do this by choosing direction $u = x - \|x\|y$, and letting $v= {u \over \|u\| }$. Then a quick calculation (using 
$\|x-\|x\|y\|^2 = 2 ( \|x\|^2- \|x\| x^Ty)$) shows that $Hx = y$.
We choose $\pm$ so that $u \neq 0$.
