Q:

Householder - Eliminating specific element of a vector

Given a column vector $$ a = \begin{bmatrix}2\\3\\4\end{bmatrix} $$
specify a Householder transformation to annihilate the third component, $a_{3}$, only.
I have been reading this description of performing this and have been coming up short over and over. 

How may I complete this problem? I am choosing ${a_{2}} = [4]$ here and I end up with $$ v = \begin{bmatrix}0\\0\\8\end{bmatrix} $$ but this is not working.
Any help is appreciated.

A:

A solution is $H=\begin{bmatrix}1&0&0\\0&0.6&0.8\\0&0.8&-0.6\end{bmatrix}$ sending vector
$a=\begin{bmatrix}2\\3\\4\end{bmatrix}$ onto vector $b=\begin{bmatrix}2\\5\\0\end{bmatrix}$.
Explanations: (The way I have proceeded is different from the way indicated by your book.)
1) The crucial step is to take an image vector $b$ that has the same norm as $a$ (because an Householder transformation is an isometry).
2) I define the difference $c=\dfrac{a-b}{\|a-b\|}$: this is clearly a direction of the symmetry which sends $a$ onto $b$. One obtains $c=\dfrac{1}{2\sqrt{5}}\begin{bmatrix}0\\-2\\4\end{bmatrix}$.
3) The matrix is given by the classical formula $H=I_3-2cc^T$.
Remark: the block structure of this matrix $H$ displays a symmetry in the $xOy$ plane, with its classical structure $\begin{bmatrix}a&b\\b&-a \end{bmatrix}$ with $a^2+b^2=1$.

