Question

Verify that the two eigenvectors in (11.8) are perpendicular, and that $\mathit{C}$ in (11.10) satisfies the condition (7.9) for an orthogonal matrix.

Short answer

The two eigenvectors are perpendicular, and that $C$ satisfies the condition for an orthogonal matrix.

Step-by-step solution

Given information

The two eigenvectors and C are show below.

$$\begin{gathered} r_1=\left(\begin{array}{c}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{array}\right) \\ {r}_2=\left(\begin{array}{c}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{array}\right) \\ C=\left(\begin{array}{c}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{array}\begin{array}{c}\frac{-2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{array}\right) \end{gathered}$$

Orthogonal matrix

An orthogonal matrix, also known as an orthonormal matrix, is a real square matrix with orthonormal vectors in its columns and rows.

Verify these eigen values are perpendicular

To verify that these eigen values are perpendicular, perform the inner product as shown below.

$$\begin{gathered} r_1^T{r}_2=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right)\left(\begin{array}{c}\frac{-2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{array}\right) \\ =\left(\frac{1}{\sqrt{5}}\right)\left(\frac{-2}{\sqrt{5}}\right)+\left(\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) \\ =\frac{-2}{5}+\frac{2}{5}=0 \end{gathered}$$

Verify that the matrix $C$satisfies the condition for orthogonality. We have,

$C^{-1}=C^T\text{or}C{C}^T=1,$

$$\begin{gathered} C{C}^T=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\\ \frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right) \\ =\left(\begin{array}{c}\frac{1}{5}+\frac{4}{5}\\ \frac{2}{5}+\frac{-2}{5}\end{array}\begin{array}{c}\frac{2}{5}+\frac{-2}{5}\\ \frac{1}{5}+\frac{4}{5}\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$