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 n is symmetric if B^T= B .

Given,

$A=\left[\begin{array}{cc}3& 3\\ 3& -5\end{array}\right]$

According to theorem 7.2.1,

$det(A-位I_n)=0$

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.

$$\begin{gathered} \det \left(A-位I_2\right)=0 \\ 鈬�det\left(\left[\begin{array}{cc}3& 3\\ 3& -5\end{array}\right]-\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]\right)=0 \\ 鈬�det\left(\left[\begin{array}{cc}3-位& 3-0\\ 3-0& -5-位\end{array}\right]\right)=0 \\ 鈬�(3-位)(-5-位)-3脳3=0 \\ 鈬�-15+2位+{位}^2-9=0 \\ 鈬�{位}^2+2位-24=0 \\ 鈬�(位-4)(位+6)=0 \\ 鈬�位=4,-6 \end{gathered}$$

The eigenvalues of the given matrix A are 4,-6.

Eigenspace for$位=4$

$$\begin{gathered} E_4=\ker \left(A-4I_2\right) \\ =ker\left(\left[\begin{array}{cc}3& 3\\ 3& -5\end{array}\right]-\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}3-4& 3-0\\ 3-0& -5-4\end{array}\right] \\ 鈬�E_4=ker\left[\begin{array}{cc}-1& 3\\ 3& -9\end{array}\right] \end{gathered}$$

Find the kernel of the matrix.

$鈬�\left[\begin{array}{cc}-1& 3\\ 3& -9\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

New second row:

localid="1668520017654" $\left(R_2+3R_1鈫�R_2\right)鈬�\left[\begin{array}{cc}-1& 3\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

From above equation we get,

$-x_1+3x_2=0鈬�-x_1=-3x_2鈬�x_1=3x_2$

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

$鈬�\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}3x_2\\ x_2\end{array}\right]=x_2\left[\begin{array}{c}3\\ 1\end{array}\right],x_2鈭�R$

Hence the eigenspace for $位=4$is $E_4=span\left[\begin{array}{c}3\\ 1\end{array}\right]$.

Eigenspace for$位=-6$

$$\begin{gathered} E_{-6}=\ker \left(A+6I_2\right) \\ 鈬�ker\left(\left[\begin{array}{cc}3& 3\\ 3& -5\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{cc}3+6& 3+0\\ 3+0& -5+6\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}9& 3\\ 3& 1\end{array}\right] \end{gathered}$$

Find the kernel of the matrix.

$鈬�\left[\begin{array}{cc}9& 3\\ 3& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

New second row:

$\left(R_2-\frac{1}{3}{R}_1鈫�R_2\right)鈬�\left[\begin{array}{cc}9& 3\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

From above equation we get,

$9x_1+3x_2=0鈬�3x_2=-9x_1鈬�x_2=-3x_1$

Therefore the eigenvector is given as

$鈬�\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_1\\ -3x_1\end{array}\right]=x_1\left[\begin{array}{c}1\\ -3\end{array}\right],x_1鈭�R$

Hence the eigenspace for $位=-6$is $E_{-6}=span\left[\begin{array}{c}1\\ -3\end{array}\right]$.

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

$$\begin{gathered} \stackrel{鈫�}{v_1}=\frac{1}{\sqrt{{\left(1\right)}^2+{(-3)}^2}}\left[\begin{array}{c}1\\ -3\end{array}\right] \\ \stackrel{鈫�}{v_1}=\frac{1}{\sqrt{10}}\left[\begin{array}{c}1\\ -3\end{array}\right] \\ \stackrel{鈫�}{v_2}=\frac{1}{\sqrt{{{\left(3\right)}^2+\left(1\right)}^2}}\left[\begin{array}{c}3\\ 1\end{array}\right] \\ \stackrel{鈫�}{v_2}=\frac{1}{\sqrt{10}}\left[\begin{array}{c}3\\ 1\end{array}\right] \end{gathered}$$

Find the inverse of orthogonal matrix:

$$\begin{gathered} S^{-1}={\left(\frac{1}{\sqrt{10}}\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right]\right)}^{-1} \\ ={\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ -\frac{3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]}^{-1} \\ =\frac{1}{{\displaystyle \frac{1}{\sqrt{10}}}脳{\displaystyle \frac{1}{\sqrt{10}}}-\left(-{\displaystyle \frac{3}{\sqrt{10}}}\right)脳{\displaystyle \frac{3}{\sqrt{10}}}}adj\left(\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]\right) \\ =\frac{1}{1}\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& -\frac{3}{\sqrt{10}}\\ \frac{3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right] \\ 鈬�S^{-1}=\frac{1}{\sqrt{10}}\left[\begin{array}{cc}1& -3\\ 3& 1\end{array}\right] \end{gathered}$$

Hence the orthogonal matrix is $S=\frac{1}{\sqrt{10}}\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right]$.

Now we have to find the diagonal matrix as below:

$$\begin{gathered} D=S^{-1}AS \\ =\frac{1}{\sqrt{10}}\left[\begin{array}{cc}1& -3\\ 3& 1\end{array}\right]\left[\begin{array}{cc}3& 3\\ 3& -5\end{array}\right]\frac{1}{\sqrt{10}}\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right] \\ =\frac{1}{\sqrt{10}}\frac{1}{\sqrt{10}}\left[\begin{array}{cc}\left(1\right)\left(3\right)+(-3)\left(3\right)& \left(1\right)\left(3\right)+(-3)(-5)\\ \left(3\right)\left(3\right)+\left(1\right)\left(3\right)& \left(3\right)\left(3\right)+\left(1\right)(-5)\end{array}\right]\left[\begin{array}{cc}1& 3\\ -3& 1\end{array}\right] \\ =\frac{1}{10}\left[\begin{array}{cc}-60& 0\\ 0& 40\end{array}\right] \\ =\left[\begin{array}{cc}\frac{-60}{10}& 0\\ 0& \frac{40}{10}\end{array}\right] \end{gathered}$$

Hence the diagonal matrix is $D=\left[\begin{array}{cc}-6& 0\\ 0& 4\end{array}\right]$.