Question 9.5.2: Apply one iteration of the QR method to the matrix that was ...

Let A^{(1)}=A be the given matrix and P_{2} represent the rotation matrix determined in Example 1. We found, using the notation introduced in the QR method, that

\begin{aligned}A_{2}^{(1)}=P_{2} A^{(1)} &=\left[\begin{array}{ccc}\frac{3 \sqrt{10}}{10} & \frac{\sqrt{10}}{10} & 0 \\-\frac{\sqrt{10}}{10} & \frac{3 \sqrt{10}}{10} & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}3 & 1 & 0 \\1 & 3 & 1 \\0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}\sqrt{10} & \frac{3}{5} \sqrt{10} & \frac{\sqrt{10}}{10} \\0 & \frac{4 \sqrt{10}}{5} & \frac{3 \sqrt{10}}{10} \\0 & 1 & 3\end{array}\right] \\& \equiv\left[\begin{array}{ccc}z_{1} & q_{1} & r_{1} \\0 & x_{2} & y_{2} \\0 & b_{3}^{(1)} & a_{3}^{(1)}\end{array}\right] .\end{aligned}

Continuing, we have

s_{3}=\frac{b_{3}^{(1)}}{\sqrt{x_{2}^{2}+\left(b_{3}^{(1)}\right)^{2}}}=0.36761 \quad \text { and } \quad c_{3}=\frac{x_{2}} {\sqrt{x_{2}^{2}+\left(b_{3}^{(1)}\right)^{2}}}=0.92998

so

\begin{aligned}R^{(1)} \equiv A_{3}^{(1)}=P_{3} A_{2}^{(1)} &=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 0.92998 & 0.36761 \\0 & -0.36761 & 0.92998\end{array}\right]\left[\begin{array}{ccc}\sqrt{10} & \frac{3}{5} \sqrt{10} & \frac{\sqrt{10}}{10} \\0 & \frac{4 \sqrt{10}}{5} & \frac{3 \sqrt{10}}{10} \\0 & 1 & 3\end{array}\right] \\&=\left[\begin{array}{ccc}\sqrt{10} & \frac{3}{5} \sqrt{10} & \frac{\sqrt{10}}{10} \\0 & 2.7203 & 1.9851 \\0 & 0 & 2.4412\end{array}\right],\end{aligned}

and

\begin{aligned}Q^{(1)}=P_{2}^{t} P_{3}^{t} &=\left[\begin{array}{ccc}\frac{3 \sqrt{10}}{10} & -\frac{\sqrt{10}}{10} & 0 \\\frac{\sqrt{10}}{10} & \frac{3 \sqrt{10}}{10} & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 0.92998 & -0.36761 \\0 & 0.36761 & 0.92998\end{array}\right] \\&=\left[\begin{array}{ccc}0.94868 & -0.29409 & 0.11625 \\0.31623 & 0.88226 & -0.34874 \\0 & 0.36761 & 0.92998\end{array}\right] .\end{aligned}

As a consequence,

\begin{aligned}A^{(2)}=R^{(1)} Q^{(1)} &=\left[\begin{array}{ccccc}\sqrt{10} & \frac{3}{5} \sqrt{10} & \frac{\sqrt{10}}{10} \\0 & 2.7203 & 1.9851 \\0 & 0 & 2.4412\end{array}\right]\left[\begin{array}{ccc}0.94868 & -0.29409 & 0.11625 \\0.31623 & 0.88226 & -0.34874 \\0 & 0.36761 & 0.92998\end{array}\right] \\&=\left[\begin{array}{ccc}3.6 & 0.86024 & 0 \\0.86024 & 3.12973 & 0.89740 \\0 & 0.89740 & 2.27027\end{array}\right]\end{aligned}

The off-diagonal elements of A^{(2)} are smaller than those of A^{(1)} by about 14%, so we have a reduction but it is not substantial. To decrease to below 0.001 we would need to perform 13 iterations of the QR method. Doing this gives

A^{(13)}=\left[\begin{array}{ccc}4.4139 & 0.01941 & 0 \\0.01941 & 3.0003 & 0.00095 \\0 & 0.00095 & 1.5858\end{array}\right]

This would give an approximate eigenvalue of 1.5858 and the remaining eigenvalues could be approximated by considering the reduced matrix

\left[\begin{array}{cc}4.4139 & 0.01941 \\0.01941 & 3.0003\end{array}\right]