Editing Householder transformation (section)

==Definition==

===Operator and transformation===

The '''Householder [[Operator (mathematics)|operator]]'''<ref>{{harvnb|Roman|2008|loc=p. 243-244}}</ref> may be defined over any finite-dimensional [[inner product space]] <math> V</math> with [[inner product]] <math> \langle \cdot, \cdot \rangle </math> and [[unit vector]] <math> u\in V</math> as
:<math> H_u(x) := x - 2\,\langle x,u \rangle\,u\,.</math><ref>{{cite book|title=Methods of Applied Mathematics for Engineers and Scientist|date=28 June 2013 |publisher=Cambridge University Press|isbn=9781107244467|pages=Section E.4.11|url=https://books.google.com/books?id=nQIlAAAAQBAJ}}</ref>

It is also common to choose a non-unit vector <math>q \in V</math>, and normalize it directly in the Householder operator's expression:<ref>{{harvnb|Roman|2008|loc=p. 244}}</ref>
:<math>H_q \left ( x \right ) = x - 2\, \frac{\langle x, q \rangle}{\langle q, q \rangle}\, q \,.</math>

Such an operator is [[Linear operator|linear]] and [[self-adjoint]].

If <math>V=\mathbb{C}^n</math>, note that the reflection hyperplane can be defined by its ''normal vector'', a [[unit vector]] <math display="inline">\vec v\in V</math> (a vector with length <math display="inline">1</math>) that is [[orthogonal]] to the hyperplane. The reflection of a [[Point (geometry)|point]] <math display="inline">x</math> about this hyperplane is the '''Householder [[linear transformation|transformation]]''':

: <math>\vec x - 2\langle \vec x, \vec v\rangle \vec v = \vec x - 2\vec v\left(\vec v^* \vec x\right), </math>

where <math>\vec x</math> is the vector from the origin to the point <math>x</math>, and <math display="inline">\vec v^*</math> is the [[conjugate transpose]] of <math display="inline">\vec v</math>.

[[File:Householdertransformation.png|thumb|The Householder transformation acting as a reflection of <math>x</math> about the hyperplane defined by <math>v</math>.]]

===Householder matrix===

The matrix constructed from this transformation can be expressed in terms of an [[outer product]] as:

: <math>P = I - 2\vec v\vec v^*</math>

is known as the '''Householder matrix''', where <math display="inline">I</math> is the [[identity matrix]].

====Properties====

The Householder matrix has the following properties:
* it is [[Hermitian matrix|Hermitian]]: <math display="inline">P = P^*</math>,
* it is [[unitary matrix|unitary]]: <math display="inline">P^{-1} = P^*</math> (via the [[Sherman-Morrison formula]]),
* hence it is [[involutory matrix|involutory]]: <math display="inline">P = P^{-1}</math>.
* A Householder matrix has eigenvalues <math display="inline">\pm 1</math>.  To see this, notice that if <math display="inline">\vec x</math> is orthogonal to the vector <math display="inline">\vec v</math> which was used to create the reflector, then <math display="inline">P_v\vec x = (I-2\vec v\vec v^*)\vec x = \vec x-2\langle\vec v,\vec x\rangle\vec v = \vec x</math>, i.e., <math display="inline">1</math> is an eigenvalue of multiplicity <math display="inline">n - 1</math>, since there are <math display="inline">n - 1</math> independent vectors orthogonal to <math display="inline">\vec v</math>.  Also, notice <math display="inline">P_v\vec v = (I-2\vec v\vec v^*)\vec v = \vec v - 2\langle\vec v,\vec v\rangle\vec v = -\vec v</math> (since <math>\vec v</math> is by definition a unit vector), and so <math display="inline">-1</math> is an eigenvalue with multiplicity <math display="inline">1</math>.
* The [[determinant]] of a Householder reflector is <math display="inline">-1</math>, since the determinant of a matrix is the product of its eigenvalues, in this case one of which is <math display="inline">-1</math> with the remainder being <math display="inline">1</math> (as in the previous point), or via the [[Matrix determinant lemma]].

====Example====

consider the normalization of a vector of 1's 

<math>\vec v=\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\1 \end{bmatrix}</math>

Then the Householder matrix corresponding to this vector is

<math>P_v=\begin{bmatrix}1&0\\0&1\end{bmatrix}-2(\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\1 \end{bmatrix})(\frac{1}{\sqrt{2}}\begin{bmatrix} 1&1 \end{bmatrix})</math>

<math>=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix} 1\\1 \end{bmatrix}\begin{bmatrix} 1&1 \end{bmatrix}</math>

<math>=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}1&1\\1&1\end{bmatrix}</math>

<math>=\begin{bmatrix}0&-1\\-1&0\end{bmatrix}</math>

Note that if we have a vector representing a coordinate in the 2D plane

<math>\begin{bmatrix}x\\y\end{bmatrix}</math>

Then in this case <math>P_v</math> flips and negates the x and y coordinates, in other words

<math>P_v\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-y\\-x\end{bmatrix}</math>

Which corresponds to reflecting the vector across the line <math>y=-x</math>, which our original vector <math>v</math> is normal to.