91Ó°ÊÓ

When a Hermitian matrix H is converted to the diagonal matrix UHU -1 , the orthonormal eigenvectors are the columns of the unitary matrixU-1.

By definition we know that

A Is symmetric$⇄A=A^T$andis orthogonal$⇄A{A}^T=I$let us test our matrix and see if it satisfies these conditions:

Let us test our matrix and see if it satisfies these conditions:

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

The transpose of A:

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

And,

$$\begin{gathered} A{A}^r=\left[\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right]\left[\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right] \\ =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]=I_4 \end{gathered}$$

We observe that these conditions are satisfied.

From Exercise 23 we proved that the only possible eigenvalues for this matrix are $Â\pm 1$. Now we can easily solve $\left(AÂ\pm I_4\right)\stackrel{→}{x}=\stackrel{→}{0}$and simultaneously prove that both $Â\pm 1$are eigenvalues of A and obtain the corresponding eigenvectors for the eigenvalues $\mathrm{λ}=1$and$\mathrm{λ}=-1$.

So,

role="math" localid="1660724562951" $$\begin{gathered} \left(AÂ\pm I_4\right)\stackrel{→}{x}=\stackrel{→}{0}\left[\left[\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x4\end{array}\right]=0 \\ \left[\begin{array}{cccc}-1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=0 \end{gathered}$$

Apply row operations role="math" localid="1660724647024" $R_4→R_4+R_1\mathrm{and}{\mathrm{R}}_3→-{\mathrm{R}}_3+{\mathrm{R}}_2$:

role="math" localid="1660724686240" $\left[\begin{array}{cccc}-1& 0& 0& 1\\ 0& -1& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=0$

Apply row operations $R_1→-R_1\mathrm{and}{\mathrm{R}}_2→-{\mathrm{R}}_2$

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

Which implies $x_1=x_4$and $x_2=x_3$, so

$\stackrel{→}{x}=\left[\begin{array}{c}{x}_4\\ x_3\\ x_3\\ x_4\end{array}\right]=x_4\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]+x_3\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]$

Therefore,

role="math" localid="1660725018335" ${\mathrm{E}}_1=\mathrm{span}\left(\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]\right)$

We actually don't need to solve $\left(A+I_4\right)\stackrel{→}{x}=0$ to find $E_{-1}$; we can use that $E_1$and $E_{-1}$are orthogonal complements, so

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

Now we can find an orthonormal eigenbasis simply by dividing the given eigenvectors

by their lengths which yields $\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ -1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ -1\\ 0\end{array}\right]$

Therefore, the orthonormal eigenbasis is given by $\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ -1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ -1\\ 0\end{array}\right]$