Question 9.4.1: Apply Householder transformations to the symmetric 4 × 4 mat...

For the first application of a Householder transformation,

\begin{aligned} &\alpha=-(1)\left(\sum_{j=2}^{4} a_{j 1}^{2}\right)^{1 / 2}=-3, r=\left(\frac{1}{2}(-3)^{2}-\frac{1}{2}(1)(-3)\right)^{1 / 2}=\sqrt{6} \\ & w =\left(0, \frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{6}\right)^{t} \end{aligned}

\begin{aligned}P^{(1)} &=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]-2\left(\frac{\sqrt{6}}{6}\right)^{2}\left[\begin{array}{r}0 \\2 \\-1 \\1\end{array}\right] \cdot(0,2,-1,1) \\&=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\0 & -\frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\0 & \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\0 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3}\end{array}\right],\end{aligned}

and

A^{(2)}=\left[\begin{array}{rrrr}4 & -3 & 0 & 0 \\-3 & \frac{10}{3} & 1 & \frac{4}{3} \\0 & 1 & \frac{5}{3} & -\frac{4}{3} \\0 & \frac{4}{3} & -\frac{4}{3} & -1\end{array}\right].

Continuing to the second iteration,

\begin{gathered}\alpha=-\frac{5}{3}, \quad r=\frac{2 \sqrt{5}}{3}, \quad w =\left(0,0,2 \sqrt{5}, \frac{\sqrt{5}}{5}\right)^{t}, \\P^{(2)}=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -\frac{3}{5} & -\frac{4}{5} \\0 & 0 & -\frac{4}{5} & \frac{3}{5}\end{array}\right],\end{gathered}

and the symmetric tridiagonal matrix is

A^{(3)}=\left[\begin{array}{rrrr}4 & -3 & 0 & 0 \\-3 & \frac{10}{3} & -\frac{5}{3} & 0 \\0 & -\frac{5}{3} & -\frac{33}{25} & \frac{68}{75} \\0 & 0 & \frac{68}{75} & \frac{149}{75}\end{array}\right]