Why is the Householder matrix orthogonal? A Householder matrix $H = I - c u u^T$, where $c$ is a constant and $u$ is a unit vector, always comes out orthogonal and full rank.

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*Why is $H$ orthogonal? I am looking for an intuitive proof rather than a rigorous one.


*How come it is full rank when $\mbox{rank} \left( u u^T \right) = \mbox{rank} (u) = 1$?

If $H$ is orthogonal, then $H^TH = I$, so let's compute that:
$$ H^TH = I - 2cuu^T + c^2 uu^Tuu^T $$
Since $u$ is a unit vector, then $u^Tu = 1$, so
$$ H^TH = I - 2cuu^T + c^2 uu^T $$
This shows that $c$ cannot be arbitrary, it must satisfy $c^2-2c = 0$ or $c = 2$ ($c=0$ is a trivial solution).
Intuitively, it represents a reflection over an $(n-1)$-dimensional hyperplane. The action of $H$ on a vector $x$ is: $Ix - 2cuu^Tx$. The first term is just $x$, while the second is the projection of $x$ on the $u$ direction, but $-2$ times of that in the $u$ direction. The latter flips the component of $x$ in the $u$ direction to the $-u$ direction.
This is clearly full rank for any unit vector $u$, since the action of $H$ on any nonzero vector is still a nonzero vector (all it does is perform a mirror operation without changing the length of the vector). Even if $u=0$, then $H$ is still full rank (corresponding to doing nothing).
