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,

$\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]$

According to theorem 7.2.1,

$\det \left(A-位I_n\right)=0$

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.

$$\begin{gathered} 鈬�det\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]-\left[\begin{array}{ccc}位& 0& 0\\ 0& 位& 0\\ 0& 0& 位\end{array}\right]\right)=0 \\ =det\left(\left[\begin{array}{ccc}0-位& 2-0& 2-0\\ 2-0& 1-位& 0-0\\ 2-0& 0-0& -1-位\end{array}\right]\right) \end{gathered}$$

$$\begin{gathered} 鈬�-位(1-位)(-1-位)-2\left(2\right(-1-位)-0)+2(0-2(1-位\left)\right)=0 \\ 鈬�\left(-位+{位}^2\right)(-1-位)-2(-2-2位)-4(1-位)=0 \\ 鈬�位+{位}^2-{位}^2-{位}^3+4+4位-4+4位=0 \\ 鈬�-{位}^3+9位=0 \\ 鈬�-位(位-3)(位+3)=0 \\ 鈬�位=0,-3,3 \end{gathered}$$

The eigenvalues of the given matrix A are 0,-3,3

Eigenspace for$位=0$

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

Now find the kernel of the matrix.

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

From the above equation we get the solution as

$$\begin{gathered} 2x_2+2x_3=0鈬�2x_2=-2x_3鈬�x_2=-x_3 \\ 2x_1+x_2=0鈬�2x_1=-x_2鈬�x_1=-\frac{1}{2}{x}_2 \\ 2x_1-x_3=0鈬�2x_1=x_3鈬�x_1=\frac{1}{2}{x}_3 \end{gathered}$$

Therefore, the eigenvector of the given matrix can be represented 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 $位=0$is $E_0=span\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.

Eigenspace for$位=-3$

$$\begin{gathered} E_{-3}=\ker \left(A+3I_3\right) \\ =ker\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]+\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{ccc}0+3& 2+0& 2+0\\ 2+0& 1+3& 0+0\\ 2+0& 0+0& -1+3\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{ccc}3& 2& 2\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right) \end{gathered}$$

Find the kernel of the matrix.

$鈬�\left(\left[\begin{array}{ccc}3& 2& 2\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Multiply first row with 1/3 to get new first row.

$\left(\frac{1}{3}脳R_1鈫�R_1\right)鈬�\left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(R_2-2R_1鈫�R_2\right)鈬�\left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& \frac{8}{3}& -\frac{4}{3}\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-2R_1鈫�R_3\right)鈬�\left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& \frac{8}{3}& -\frac{4}{3}\\ 0& -\frac{4}{3}& \frac{2}{3}\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New second row:

$\left(\frac{3}{8}脳R_2鈫�R_2\right)鈬�\left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& 1& -\frac{1}{2}\\ 0& -\frac{4}{3}& \frac{2}{3}\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New third row:

$\left(R_3+\frac{4}{3}脳R_2鈫�R_3\right)鈬�\left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New first row:

$B^T=B\left(R_1-\frac{2}{3}脳R_2鈫�R_1\right)鈬�\left(\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\left(1\right)$

From equation (1) we get,

$$\begin{gathered} x_1+x_3=0鈬�x_1=-x_3 \\ {x}_2-\frac{1}{2}{x}_3=0鈬�x_2=\frac{1}{2}{x}_3 \end{gathered}$$

Therefore the eigenvector is given as

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

Hence the eigenspace for $位=-3$is $鈬�E_{-3}=span\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$.

Eigenspace for$位=3$

$$\begin{gathered} E_3=\ker \left(A-3I_3\right) \\ =ker\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]-\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\right) \\ =ker\left[\begin{array}{ccc}-3& 2& 2\\ 2& -2& 0\\ 2& 0& -4\end{array}\right] \end{gathered}$$

Finding the kernel:

$鈬�\left[\begin{array}{ccc}-3& 2& 2\\ 2& -2& 0\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row:

$\left(-\frac{1}{3}脳R_1鈫�R_1\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 2& -2& 0\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(R_2-2R_1鈫�R_2\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& -\frac{2}{3}& \frac{4}{3}\\ 2& 0& -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]$$\left(-\frac{1}{3}脳R_1鈫�R_1\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 2& -2& 0\\ 2& 0& -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]$

To get new third row:

$\left(R_3-2R_1鈫�R_3\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& -\frac{2}{3}& \frac{4}{3}\\ 0& \frac{4}{3}& -\frac{8}{3}\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(-\frac{3}{2}脳R_2鈫�R_2\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& 1& -2\\ 0& \frac{4}{3}& -\frac{8}{3}\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-\frac{4}{3}脳R_2鈫�R_3\right)鈬�\left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row:

$\left(R_1+\frac{2}{3}脳R_2鈫�R_1\right)鈬�\left[\begin{array}{ccc}1& 0& -2\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\left(2\right)$

From the above equation we get,

$$\begin{gathered} x_1-2x_3=0鈬�x_1=2x_3 \\ {x}_2-2x_3=0鈬�x_2=2x_3 \end{gathered}$$

The eigenvectors are:

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

Hence the eigenspace for$位=3$is$E_3=span\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$

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

$$\begin{gathered} \frac{1}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+2^2}}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]鈬�\frac{1}{3}\left[\begin{array}{c}1\\ 2\\ 2\end{array}\right] \\ \frac{1}{\sqrt{{(-2)}^2+{\left(1\right)}^2+2^2}}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]鈬�\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ \frac{1}{\sqrt{2^2+2^2+1^2}}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]鈬�\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \end{gathered}$$

Hence the orthonormal eigenbasis are $\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$.