91影视

• An orthogonal matrix, also known as an orthonormal matrix, is a real square matrix whose columns and rows are orthonormal vectors in linear algebra.
• Where ${\mathit{Q}}^{\mathbf{T}}$ is the transpose of Qand Iis the identity matrix is one approach to describe this.

By definition we know that

A Is symmetric $鈬�A=A^T$ and A is orthogonal $鈬�A{A}^T$let us test our matrix and see if it satisfies these conditions:

$A=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]$

The transpose of A :

$A^T=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]=A$

And,

$$\begin{gathered} A{A}^T=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right] \\ =\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]=I_5 \end{gathered}$$

We observe that these conditions are satisfied.

From Exercise 23, we found out that the only possible eigenvalues for this matrix are $卤1$. Now we can easily solve $\left(A卤I_5\right)\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$, and simultaneously prove that both $卤1$ are eigenvalues of A and obtain the corresponding eigenvectors for the eigenvalues $位=1$and $位=-1$.

CASE : When $位=1$

localid="1660732828960" $$\begin{gathered} \left(A卤I_5\right)\stackrel{鈫�}{x}=\stackrel{鈫�}{0}鈫�\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0 \\ \left[\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 0& -1& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& -1& 0\\ 0& 0& 0& 0& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0 \end{gathered}$$

Apply row operations $R_5鈫�R_5+R_1$and $R_4鈫�R_4+R_2$

$\left[\begin{array}{ccccc}-1& 0& 0& 0& 1\\ 0& -1& 0& 1& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0$

Which implies $x_1=x_5$and $x_2=x_4$,

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=x_5\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 1\end{array}\right]+x_4\left[\begin{array}{c}0\\ 1\\ 0\\ 1\\ 0\end{array}\right]+x_3\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]$

Therefore,

$E_1=span\left(\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right]\right)$

CASE : When $位=-1$

$$\begin{gathered} \left(A卤I\right)x=0鈫�\left[\left[\begin{array}{ccccc}0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\end{array}\right]+\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0 \\ \left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 1& 0& 1& 0\\ 0& 0& 2& 0& 0\\ 0& 1& 0& 1& 0\\ 1& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0 \end{gathered}$$

Apply row operations $R_5鈫�R_5-R_1$and $R_4鈫�R_4-R_2$

$\left[\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 1& 0& 1& 0\\ 0& 0& 2& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=0$

Which implies $x_1=-x_5,x_3=0$and $x_2=-x_4$,

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ 0\\ -x_2\\ -x_1\end{array}\right]=x_1\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ -1\end{array}\right]+x_2\left[\begin{array}{c}0\\ 1\\ 0\\ -1\\ 0\end{array}\right]$

Therefore,

$E_{-1}=span\left(\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ -1\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ -1\\ 0\end{array}\right]\right)$

Now we can find an orthonormal eigenbasis first by simply dividing the given eigenvectors by their lengths, which yields

$\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 0\\ 1\\ 0\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ -1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\\ 0\end{array}\right]$

So one possible choice for the orthogonal matrix is

$S=\frac{1}{\sqrt{2}}\left[\begin{array}{ccccc}1& 0& 0& 1& 0\\ 0& 1& 0& 0& 1\\ 0& 0& \sqrt{2}& 0& 0\\ 0& 1& 0& 0& -1\\ 1& 0& 0& -1& 0\end{array}\right]$

Therefore, the orthogonal matrix is$S=\frac{1}{\sqrt{2}}\left[\begin{array}{ccccc}1& 0& 0& 1& 0\\ 0& 1& 0& 0& 1\\ 0& 0& \sqrt{2}& 0& 0\\ 0& 1& 0& 0& -1\\ 1& 0& 0& -1& 0\end{array}\right]$