91影视

Theorem 7.2.1

Consider an n x n matrix A and a scalar $位$. Then $位$is an eigenvalue of A if (and only if)

$\det \left(A-位I_n\right)=0$

This is called the characteristic equation (or the secular equation) of matrix A.

Definition 7.3.1

Consider an eigenvalue $位$of an n x n matrix A. Then the kernel of the matrix $A-位I_n$is called the Eigenspace associated with $位$, denoted by $E_{位}$:

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

Now consider the given matrix as represented below. We have to find an orthonormal eigenbasis of the matrix A.

$A=\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]$

Now by Theorem 7.2.1, solving the characteristic equation of the given matrix A as represented below.

$$\begin{gathered} \det \left(A-位I_2\right)=0 \\ 鈬�det\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]\right)=0 \\ 鈬�det\left(\left[\begin{array}{cc}6-位& 2\\ 2& 3-位\end{array}\right]\right)=0 \\ 鈬�(6-位)(3-位)-2脳2=0 \\ 鈬�{位}^2-9位+14=0 \\ 鈬�(位-7)(位-2)=0 \\ 鈬�位=2,7 \end{gathered}$$

Therefore the eigenvalues of the given matrix A are 2 and 7.

Now from Definition 7.3 .1 for eigenvalue $位=2$we get the eigenspaces as follows.

$$\begin{gathered} E_2=ker(A-2I_2) \\ =ker\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}6-2& 2-0\\ 2-0& 3-2\end{array}\right] \\ {E}_2=ker\left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix (A-2I₂) as represented below.

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

Now subtracting 1/2 times the 1^st row from the 2^nd row to get a new 2^nd row as represented below.

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

From the above equation we get the solution as represented below.

$4x_1+2x_2=0鈬�2x_1+x_2=0鈬�x_2=-2x_1$

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $位=2$can be represented as shown below.

$鈬�\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_1\\ -2x_1\end{array}\right]=x_1\left[\begin{array}{c}1\\ -2\end{array}\right],x_1鈭�R$

Therefore

$鈬�E_2=span\left[\begin{array}{c}1\\ -2\end{array}\right]$

Eigenvalue $位=7$ we get the eigenspaces as follows.

$$\begin{gathered} E_7=ker(A-7I_2) \\ =ker\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}6-7& 2-0\\ 2-0& 3-7\end{array}\right] \\ {E}_7=ker\left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix (A-7I₂) as represented below.

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

From the above equation we get the solution as represented below.

$-1x_1+2x_2=0鈬�x_1=2x_2$

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $位=7$can be represented as shown below.

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

Therefore

$鈬�E_7=span\left[\begin{array}{c}2\\ 1\end{array}\right]$

This represents of the eigenspace graph as follows.

We can see that the Eigenspaces E₂ and E₇ are perpendicular to each other. Therefore we can find an orthonormal eigenbasis of the symmetric matrix A by dividing the eigenvectors by their magnitudes as represented below.

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

$\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ -2\end{array}\right]and\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$

Therefore, the orthonormal eigenbasis of symmetric matrix is$\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ -2\end{array}\right]and\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$