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.

$A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]$

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

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

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

Therefore the eigenvalues of the given matrix A are -1,0 and 2.

Now from Definition 7.3.1 for eigenvalue $位=-1$we get the eigenspaces as follows.

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

Now finding the kernel of the matrix (A+1I₃) as represented below.

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

Now subtracting the 2^nd row from the 3^rd row to get a new 3^rd row as represented below.

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

Now subtracting the 1^st row from the 3^rd row to get a new 3^rd row as represented below.

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

From equation (1) we get the solutions as represented below.

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

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $位=-1$can be represented as shown below.

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

Therefore

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

Similarly from Definition 7.3.1 for eigenvalue $位=0$we get the eigenspaces as follows.

$$\begin{gathered} E_0=ker(A-0I_3) \\ {E}_0=ker\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix $(A-0I_3)$as represented below.

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

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

$$\begin{gathered} 0x_1+0x_2+x_3=0鈬�x_3=0 \\ {x}_1+x_2+x_3=0鈬�x_1+x_2+0=0鈬�x_1=-x_2 \end{gathered}$$

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $位=0$can be represented as shown below.

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

Therefore

$鈬�E_0=span\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right]$

Similarly from Definition 7.3.1 for eigenvalue $位=2$ we get the eigenspaces as follows.

$$\begin{gathered} E_2=ker(A-2I_3) \\ =ker\left(\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]-\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\right) \\ =ker\left[\begin{array}{ccc}0-2& 0& 1-0\\ 0& 0-2& 1-0\\ 1-0& 1-0& 1-2\end{array}\right] \\ =ker\left[\begin{array}{ccc}-2& 0& 1\\ 0& -2& 1\\ 1& 1& -1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix $(A-2I_3)$as represented below.

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

Now adding 1/2 times the 1^st row to the 3^rd row to get a new 3^rd row as represented below.

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

Now adding 1/2 times the 2^nd row to the 3^rd row to get a new 3^rd row as represented below.

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

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

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

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

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

Therefore

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

The graph represent the eigenspaces and as shown below.

We can see that the eigenspaces E_-1, E₀ and E₂ are perpendicular to each other. Therefore we can find an orthonormal eigenbasis of the symmetric matrix A by dividing the eigenvectors by their lengths as represented below.

$$\begin{gathered} \frac{1}{\sqrt{{(-1)}^2+{(-1)}^2+1^2}}\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]鈬�\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right] \\ \frac{1}{\sqrt{{(-1)}^2+{\left(1\right)}^2+\left(\right)}}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]鈬�\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right] \\ \frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]鈬�\frac{1}{\sqrt{6}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right] \end{gathered}$$

Therefore, the orthonormal Eigenbasis of symmetric matrix is

$\frac{1}{\sqrt{3}}\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],\frac{1}{\sqrt{6}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$