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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,

$A=\left[\begin{array}{cc}3& 2\\ 2& 3\end{array}\right]$

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.

$$\begin{gathered} \det \left(A-位I_2\right)=0 \\ =det\left(\left[\begin{array}{cc}3& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]\right)=0 \\ 鈬�det\left(\left[\begin{array}{cc}3-位& 2-0\\ 2-0& 3-位\end{array}\right]\right)=0 \end{gathered}$$

$$\begin{gathered} 鈬�(3-位)(3-位)-2脳2=0 \\ 鈬�{位}^2-6位+5=0 \\ 鈬�(位-5)(位-1)=0 \\ 鈬�位=5,1 \end{gathered}$$

The eigenvalues of the given matrix A are 5,1.

Eigenspace for$位=5$

$$\begin{gathered} E_5=\ker \left(A-5I_2\right) \\ =ker\left(\left[\begin{array}{cc}3& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{cc}3-5& 2-0\\ 2-0& 3-5\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right] \end{gathered}$$

Find the kernel of the matrix.

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

New second row:

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

From above equation we get,

$-2x_1+2x_2=0鈬�-2x_1=-2x_2鈬�x_1=x_2$

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

$鈬�\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_2\\ x_2\end{array}\right]=x_2\left[\begin{array}{c}1\\ 1\end{array}\right],x_2鈭�R$

Hence the eigenspace for $位=5$is $E_5=span\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Eigenspace for$位=1$$位=1$

width="164">E1=kerA-I2=ker3223-1001=ker2222

Find the kernel of the matrix.

$=\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

New second row:

$\left(R_2-R_1鈫�R_2\right)=\left[\begin{array}{cc}2& 2\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

From above equation we get,

$2x_1+2x_2=0鈬�2x_1=-2x_2鈬�x_1=-x_2$

Therefore the eigenvector is given as

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

Hence the eigenspace for $位=1$is $E_1=span\left[\begin{array}{c}-1\\ 1\end{array}\right]$.