Let $x=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$
I want to use a Householder reflector U to keep only first element in vector x, and make everything else zero
but I'm doing something wrong...
$U=I-\frac{uu^T}{\beta}$
$\beta=\frac{\left \| u \right \|_2^2}{2}$
$Ux=x-u$
$\beta=\frac{16}{2}=8$
$u=\begin{bmatrix} 0\\ 2\\ 3 \end{bmatrix}$
$U=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} -\frac{1}{8} \begin{bmatrix} 0 & 0 & 0\\ 0 & 4 & 6\\ 0 & 6 & 9 \end{bmatrix}$
$U=\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{2} & -\frac{3}{4}\\ 0 & -\frac{3}{4} & -\frac{1}{8} \end{bmatrix}$
but $Ux \neq x-u$ (will not make anything zero)
Please tell me what I'm doing wrong
EDIT:
If $u=\begin{bmatrix} -3\\ 2\\ 3 \end{bmatrix}$
then $uu^T=\begin{bmatrix} 9 & -6 & -9 \\ -6 & 4 & 6 \\ -9 & 6 & 9 \end{bmatrix}$
but it's still not working
Your $u$ should be calculated as $u=x-\|x\|(1,0,0)^T$, but you mistook it as $u=x-(1,0,0)^T$.