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Householder matrix

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Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Problem: Let $H$ be an $n \times n$ householder matrix given by \(\displaystyle H = I_n-2 \frac{vv^T}{v^Tv}\) for any non-zero $n$-length column vector $v$. Show that $H=H^T$ and $HH^T=I_n$.

Attempt: a) Let's look at $H$ written out in matrix form for an 3x3 matrix. I believe it would look like this.

\(\displaystyle H_{3 \times 3}=\left( \begin{array}{ccc}1-2 \frac{vv^T}{v^Tv} &-2 \frac{vv^T}{v^Tv}&-2 \frac{vv^T}{v^Tv} \\ -2 \frac{vv^T}{v^Tv}&1-2 \frac{vv^T}{v^Tv}&-2 \frac{vv^T}{v^Tv}\\-2 \frac{vv^T}{v^Tv}&-2 \frac{vv^T}{v^Tv}&1-2 \frac{vv^T}{v^Tv} \end{array} \right)\).

Clearly here $H=H^T$. I believe showing this in the general case requires some tricky notation for me. Any ideas to get started on generalizing this?

b) To show $HH^T=I_n$ we start by looking at \(\displaystyle \left(I_n-2 \frac{vv^T}{v^Tv}\right) \left(I_n-2 \frac{vv^T}{v^Tv} \right)^T\)

I'm familiar with rules of matrix multiplication and transpose when products are involved, but when sums are there I'm hoping I can FOIL these terms.

Let's start with the fact from (a) that I haven't proved yet that $H=H^T$. That means that:

\(\displaystyle \left(I_n-2 \frac{vv^T}{v^Tv}\right)^T=\left(I_n-2 \frac{vv^T}{v^Tv}\right)\) so the original product becomes:

\(\displaystyle \left(I_n-2 \frac{vv^T}{v^Tv}\right) \left(I_n-2 \frac{vv^T}{v^Tv}\right) \)

I don't know exactly where to go from here...
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Clearly here $H=H^T$. I believe showing this in the general case requires some tricky notation for me. Any ideas to get started on generalizing this?
Using well known properties of tranposition and taking into account that $\dfrac{2}{v^tv}$ is an scalar:

$H ^T= \left(I_n-2 \dfrac{vv^T}{v^Tv}\right)^T=I_n^T-\dfrac{2}{v^tv}(vv^T)^T=I_n^T-\dfrac{2}{v^tv}(v^T)^Tv^T\\=I_n^T-\dfrac{2}{v^tv}(vv^T)=I_n-2 \dfrac{vv^T}{v^Tv}=H$

To show $HH^T=I_n$ use that $HH^T=H^2,$ that $I_n$ commutes with all matrix $n\times n$ and
$(A-B)^2=A^2-2AB+B^2$ if $AB=BA.$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here you have a complete proof of $HH^T=I_n$ for if you want to verify your work:

Using that $I_n$ commutes with all $n\times n$ matrix:

$HH^T=H^2= \left( I_n-2 \dfrac{vv^T}{v^Tv}\right)^2=I_n^2-4I_n \dfrac{vv^T}{v^Tv}+4\left (\dfrac{vv^T}{v^Tv} \right)^2\\=I_n-4\dfrac{vv^T}{v^Tv}+4\dfrac{vv^T}{v^Tv}\dfrac{vv^T}{v^Tv}$

Using that $v^Tv$ is an scalar:

$\dfrac{vv^T}{v^Tv}\dfrac{vv^T}{v^Tv}=\dfrac{1}{v^Tv}\dfrac{1}{v^Tv}v(v^Tv)v^T=\dfrac{1}{v^Tv}\dfrac{1}{v^Tv}(v^Tv)(vv^T)=\dfrac{vv^T}{v^Tv}$

So, $HH^T=I_n,$ i.e. $H$ is orthogonal.