Question 9.5.1: Find a rotation matrix P with the property that PA has a zer...
Find a rotation matrix P with the property that PA has a zero entry in the second row and first column, where
A=\left[\begin{array}{lll} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{array}\right] .
The form of P is
P=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\-\sin \theta & \cos \theta & 0 \\0 & 0 & 1\end{array}\right] \quad \operatorname{so} P A=\left[\begin{array}{ccc}3 \cos \theta+\sin \theta & \cos \theta+3 \sin \theta & \sin \theta \\-3 \sin \theta+\cos \theta & -\sin \theta+3 \cos \theta & \cos \theta \\0 & 1 & 3\end{array}\right].
The angle θ is chosen so that −3 sin θ + cos θ = 0, that is, so that \tan \theta=\frac{1}{3} . Hence
\cos \theta=\frac{3 \sqrt{10}}{10} . \quad \sin \theta=\frac{\sqrt{10}}{10}
and
P A=\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}3 & 1 & 0 \\1 & 3 & 1 \\0 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}\sqrt{10} & \frac{3}{5} \sqrt{10} & \frac{1}{10} \sqrt{10} \\0 & \frac{4}{5} \sqrt{10} & \frac{3}{10} \sqrt{10} \\0 & 1 & 3\end{array}\right].
Note that the resulting matrix is neither symmetric nor tridiagonal.