Orthonormal basis using Householder matrices

Question

I have read about how to find a QR factorization of a matrix using Householder matrices. I have also read that it is possible to find an orthonormal basis with them, but I can't figure out how I should do that. Thanks for the help.

Answer 1

Here is the idea: Given a matrix $A=(v_1, \dots, v_n) \in \mathbb{R}^{n \times n}$ with column vectors $v_j$ you want to find $n-1$ Householder matrices such that $Q_{n-1} \cdots Q_1A=R$. The best way to understand how to find these matrices is to look at the problem geometrically: A Householder matrix is a reflection about a hyperplane. Start with $v_1$. Costruct a Householder matrix $Q_1$ in such a way that $$Q_1v_1 = (r_1,0,0,\dots,0), \ r_1 \in \mathbb{R}$$. So we want to reflect the vector $v_1$ in such a way that the resulting vector lies on the $x_1$-axis. When you multiply the whole matrix $A$ with $Q_1$ you get:

$$ Q_1A=\begin{pmatrix} r_1 & * & \dots & * \\ 0 & * & \dots & * \\ \vdots & \vdots & \ddots & * \\ 0 & * & \dots & * \end{pmatrix} := B = (b_1,b_2, \dots,b_n)$$

Now consider the vector $b_2$ without the first entry. Call this vector $\overline{b_2}$.This is now a vector in $\mathbb{R}^{n-1}$. Again we search for a Householder matrix $Q$ but now with $Q \in \mathbb{R}^{n-1 \times n-1}$ such that

$$Q\overline{b_2} = (r_2,0,0,\dots,0) \in \mathbb{R}^{n-1 \times n-1}$$

Now define

$$Q_2 = \begin{pmatrix} 1 & 0 \\ 0 & Q \end{pmatrix}$$

When you multiply this matrix with $Q_1A$, you don't act on the first line and first collumn of $Q_1A$. So the nice property of the first column of $Q_1A$ is preserved. Hence

$$ Q_2Q_1A=\begin{pmatrix} r_1 & * & * & \dots & * \\ 0 & r_2 &* & \dots & * \\ 0 & 0 & *& \dots & * \\ \vdots & \vdots & \vdots & \ddots & * \\ 0 & 0 & *& \dots & * \end{pmatrix}$$

Continue in this way to get $Q_{n-1} \cdots Q_1A=R$.