Exercise 2.1.7

When $n=3$, we have $A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$, For the QR algorithm, we use $Q_3$ found in problem 5. That is: $A_1 = Q^{-1}_3AQ_3 = Q^T_3AQ_3 =

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ $\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=

$\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$

We see that the new matrix $A_1$ is still tridiagonal, with the same eigenvalues as $A$.