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- Jan 26, 2012

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**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...