Why $Hx=x-(\rho u^Tx)u$? A Householder reflection is  a matrix of the form
$$H=I-\rho uu^T$$
with $\rho=2/\|u\|^2$.
Obviously, $Hx=x-\rho uu^Tx$. 
Textbook http://www.mathworks.se/moler/leastsquares.pdf says that 
$$Hx=x-(\rho u^Tx)u$$
However, I can not see the proof of this statement.
 A: Note that $u^t x \in \mathbb R$ is a scalar, and scalar multipliation $(\lambda,u) \mapsto \lambda u$ is usually written on the left, so 
$$ Hx = x - \rho u u^tx = x - (\rho u^t x)u. $$
A: Your question boils down to: why is $(uu^T)x = (u^Tx)u$? There are actually four reasons:


*

*Matrix multiplication is associative. Hence $(uu^T)x=u(u^Tx)$.

*Note that $u^Tx$ is a $1\times1$ matrix whose only entry is the scalar $s:=\langle u,x\rangle=\sum x_iu_i$. So, $u^Tx=[s]$, where the pair of square brackets is used to indicate that $u^Tx$ is a matrix.

*Now comes a point that is both subtle and obvious: $u[s]=su$. Note that the LHS is a matrix product (an $n\times1$ matrix multiplied by a $1\times1$ matrix) while the RHS is a scalar multiplication (the vector $u$ is scaled by the scalar $s$). This equality is perhaps so obvious that it is taken for granted and isn't discussed in most textbooks.

*Therefore $(uu^T)x=u(u^Tx)=u[s]=su$. Now, if we have identify $s$ with $[s]$, we can rewrite $su$ as $(u^Tx)u$. Hence $(uu^T)x=(u^Tx)u$.

A: because 
$$(uu^T)x = u(u^Tx)$$
$u^Tx$ is not a vector.
so
$u(u^Tx) = (u^Tx)u$
