91Ó°ÊÓ

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

Let$A=B^3$ , determine the eigenvalues of the matrix A

$$\begin{gathered} det\left(A-\mathrm{λ}{l}_n\right)=0 \\ \left[\begin{array}{cc}\frac{12}{5}-\mathrm{λ}& \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-\mathrm{λ}\end{array}\right]=0 \\ \left(\frac{12}{5}-\mathrm{λ}\right)\left(\frac{33}{5}-\mathrm{λ}\right)-\left(\frac{14}{5}\right)\left(\frac{14}{5}\right)=0 \\ 25{\mathrm{λ}}^2-225\mathrm{λ}+200=0 \\ \left(\mathrm{λ}-8\right)\left(\mathrm{λ}-1\right)=0 \end{gathered}$$

The eigenvalues are $\mathrm{λ}=1$, and $\mathrm{λ}=8$. Now we obtain the eigenvectors.

CASE 1: When$\mathrm{λ}=1$

$$\begin{gathered} \left[A-l\right]\tilde{x}=0⇒\left[\begin{array}{cc}\frac{12}{5}-1& \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0 \\ \left[\begin{array}{cc}\frac{7}{5}& \frac{14}{5}\\ \frac{14}{5}& \frac{28}{5}\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0 \end{gathered}$$

apply row operation$R_2→2R_1-R_2$:

$\left[\begin{array}{cc}\frac{7}{5}& \frac{14}{5}\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

apply row operation$R_1→5R_1$

$\left[\begin{array}{cc}7& 14\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

so here we have$a=-2b$ , choosing$b=1$yields$a=-2$ . The eigenvector corresponding to the eigenvalue$\mathrm{λ}=1$is

$\stackrel{→}{v}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$

therefore,

$E_1=span\left(\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]\right)$

CASE 2: When$\mathrm{λ}=8$

role="math" localid="1660104429366" $$\begin{gathered} \left[A-l\right]\tilde{x}=0→\left[\begin{array}{cc}\frac{12}{5}-8& \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-8\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0 \\ \left[\begin{array}{cc}-\frac{28}{5}& \frac{14}{5}\\ \frac{14}{5}& -\frac{7}{5}\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0 \end{gathered}$$

apply row operation$R_2→2R_2+R_1$ and$R_1→5R_1$:

$\left[\begin{array}{cc}-28& 14\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

so here we have $a=\frac{1}{2}b$, choosing b = 2 yields a = 1 . The eigenvector corresponding to the eigenvalue $\mathrm{λ}=8$is

${\stackrel{→}{v}}_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$

therefore,

$E_8=span\left(\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]\right)$

In decomposition $A=SD{S}^{-1}$, the orthogonal matrix is given by

$S=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]$

and the diagonal matrix is given by

$D=\left[\begin{array}{cc}8& 0\\ 0& 1\end{array}\right]$

Now let

$P=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]$

notice that $P^3=D$so ${\left(SD{S}^{-1}\right)}^3=S{P}^3{S}^{-1}=A$. So, we compute B such that

$$\begin{gathered} S^{-1}BS=P→B=SP{S}^{-1} \\ =\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{c}1\\ \sqrt{5}\end{array}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\right]}^{-1} \\ =\frac{1}{\sqrt{5}}\left[\begin{array}{cc}2& -2\\ 4& 1\end{array}\right]\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right] \\ =\frac{1}{5}\left[\begin{array}{cc}6& 2\\ 2& 9\end{array}\right] \end{gathered}$$