91影视

• In linear algebra, a symmetric matrix is a square matrix that does not change when its transpose is calculated.
• A symmetric matrix is defined as one whose transpose is identical to the matrix itself.
• A square matrix of size n x nis symmetric if B^T=B.

Given,

$A=\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]$

According to theorem 7.2.1,

role="math" localid="1668575473197" $\det \left(A-位I_n\right)=0$

The formula for finding eigenspace is

$E_{位}=\ker \left(A-位I_n\right)=\left\{\stackrel{鈫�}{v}\text{in}{\mathrm{鈩�}}^n:A\stackrel{炉}{v}=位\stackrel{炉}{v}\right\}$

Now we need to find an orthonormal eigenbasis for the given matrix A.

$$\begin{gathered} 鈬�det\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}位& 0& 0\\ 0& 位& 0\\ 0& 0& 位\end{array}\right]\right)=0 \\ 鈬�det\left(\left[\begin{array}{ccc}1-位& -2& 2\\ -2& 4-位& -4\\ 2& -4& 4-位\end{array}\right]\right)=0 \\ 鈬�(1-位)\left[{(4-位)}^2-16\right]+2[-2(4-位)+8]+2[8-2(4-位\left)\right]=0 \\ 鈬�(1-位)\left[16+{位}^2-8位-16\right]+2[-8+2位+8]+2[8-8+2位]=0 \\ 鈬�{位}^2-8位-{位}^3+8{位}^2+4位+4位=0 \\ 鈬�-{位}^3+9{位}^2=0 \\ 鈬�-{位}^2(位-9)=0 \\ 鈬�位=0,0,9 \end{gathered}$$

The eigenvalues of the given matrix A are 0,0,9.

Eigenspace for$位=0$

$$\begin{gathered} E_0=\ker \left(A-0I_3\right) \\ =ker\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\right)=0 \\ =ker\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right] \end{gathered}$$

Now find the kernel of the matrix.

$鈬�\left[\begin{array}{ccc}1& -2& 2\\ -2& 1& -4\\ 2& -4& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-2R_1鈫�R_3\right)鈬�\left[\begin{array}{ccc}1& -2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New second row:

$\left(R_2+2R_1鈫�R_2\right)鈬�\left[\begin{array}{ccc}1& -2& 2\\ 0& 0& 0\\ 2& -4& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

From the above equation we get the solution as

$$\begin{gathered} x_1-2x_2+2x_3=0 \\ 鈬�x_1=2x_2-2x_3 \end{gathered}$$

Therefore, the eigenvector of the given matrix can be represented as:

$$\begin{gathered} 鈬�\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2x_2-2x_3\\ x_2\\ x_3\end{array}\right] \\ =\left[\begin{array}{c}2x_2\\ x_2\\ 0\end{array}\right]+\left[\begin{array}{c}-2x_3\\ 0\\ x_3\end{array}\right] \\ =x_2\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]-x_3\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right],x_2,x_3鈭�R \end{gathered}$$

Hence the eigenspace for $位=0$is $E_0=span\left(\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]\right)$.

Eigenspace for$位=9$

$$\begin{gathered} E_0=\ker \left(A-9I_3\right) \\ =ker\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\right)=0 \\ =ker\left[\begin{array}{ccc}-8& -2& 2\\ -2& -5& -4\\ 2& -4& -5\end{array}\right] \end{gathered}$$

Find the kernel of the matrix.

$=ker\left[\begin{array}{ccc}-8& -2& 2\\ -2& -5& -4\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row.

role="math" localid="1668579108489" $$\begin{gathered} \left(-\frac{1}{8}脳R_1鈫�R_1\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ -2& -5& -4\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(R_1-\frac{1}{4}{R}_2鈫�R_1\right)=ker\left[\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

To get new third row:

$$\begin{gathered} \left(R_3-2R_1鈫�R_3\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& -\frac{9}{2}& -\frac{9}{2}\\ 0& \frac{9}{2}& -\frac{9}{2}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(R_3+\frac{9}{2}{R}_2鈫�R_3\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

New second row:

$$\begin{gathered} \left(R_2+2R_1鈫�R_2\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& -\frac{9}{2}& -\frac{9}{2}\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(\frac{1}{5}脳R_2鈫�R_2\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& 1& 1\\ 0& -\frac{9}{2}& -\frac{9}{2}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

From equation above we get,

$$\begin{gathered} x_1+0-\frac{1}{2}{x}_3=0鈬�x_1=\frac{1}{2}{x}_3 \\ 0+x_2+x_3=0 \end{gathered}$$

Therefore the eigenvector is given as

$鈬�\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{x}_3\\ -x_3\\ x_3\end{array}\right]=\frac{1}{2}{x}_3\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],x_3鈭�R$

Hence the eigenspace for $位=9$is $E_9=span\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.

The eigenspaces are perpendicular to each other. The orthonormal eigenbasis can be calculated as follows:

$$\begin{gathered} \stackrel{鈫�}{v_1}=\frac{1}{\sqrt{2^2+1^2+0^2}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right] \\ \stackrel{鈫�}{v_1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right] \\ \stackrel{鈫�}{v_2}=\frac{\left(\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]-\left({\displaystyle \frac{1}{\sqrt{5}}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right].\left[\begin{array}{c}2\\ 0\\ -1\end{array}\right]\right)\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]\right)}{\left|\left(\left[\begin{array}{c}2\\ -1\end{array}\right]-\left[\begin{array}{c}2\\ \sqrt{5}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right)\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]\right|} \end{gathered}$$

$$\begin{gathered} =\frac{\sqrt{5}}{3}\left[\begin{array}{c}\frac{2}{5}\\ -\frac{4}{5}\\ -1\end{array}\right] \\ =\stackrel{鈫�}{v_2}\left[\begin{array}{c}\frac{2}{3\sqrt{5}}\\ -\frac{4}{3\sqrt{5}}\\ \stackrel{鈫�}{v_3}\end{array}\right] \end{gathered}$$

$$\begin{gathered} \stackrel{鈫�}{v_3}=\frac{5}{3}\left[-\frac{1}{3\sqrt{5}}\right]\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right] \\ S=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \end{gathered}$$

Find the inverse of orthogonal matrix:

$$\begin{gathered} S^{-1}={\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right]}^{-1} \\ =\frac{1}{\left({\displaystyle \frac{20}{27}}\right)}adj\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \\ 鈬�S^{-1}=S=\left[\begin{array}{ccc}-\frac{27}{10\sqrt{5}}& -\frac{9}{10\sqrt{5}}& -\frac{9}{20}\\ -\frac{27}{20\sqrt{5}}& \frac{9}{5\sqrt{5}}& \frac{9}{10}\\ 0& \frac{9\sqrt{5}}{20}& -\frac{9}{10}\end{array}\right] \end{gathered}$$

Hence the orthogonal matrix is $S=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right]$.

Now we have to find the diagonal matrix as below:

$$\begin{gathered} D=S^{-1}AS \\ \left[\begin{array}{ccc}-\frac{27}{10\sqrt{5}}& -\frac{9}{10\sqrt{5}}& -\frac{9}{20}\\ -\frac{27}{20\sqrt{5}}& \frac{9}{5\sqrt{5}}& \frac{9}{10}\\ 0& \frac{9\sqrt{5}}{20}& -\frac{9}{10}\end{array}\right]\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \end{gathered}$$

Hence the diagonal matrix is $D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 9\end{array}\right]$.