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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

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1 Answer 1

up vote 0 down vote accepted

Your $u$ should be calculated as $u=x-\|x\|(1,0,0)^T$, but you mistook it as $u=x-(1,0,0)^T$.

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