POWER METHODS Find all eigenvalues and eigenvectors of the following matrix using the power, shifted inverse power, and shifted power methods: [latex] \mathbf{A}=\left[\begin{array}{cccc} 4 & -3 & 3 & -9 \\ -3 & 6 & -3 & 11 \\ 0 & 8 & -5 & 8 \\ 3 & -3 & 3 & -8 \end{array}\right], \quad \mathbf{x}_{1}=\left\{\begin{array}{l} 0 \\ 1 \\ 0 \\ 1 \end{array}\right\} [/latex]

Apply the user-defined function PowerMethod to find the dominant eigenvalue of A:

Question 9.3: POWER METHODS Find all eigenvalues and eigenvectors of the f...

Numerical Methods for Engineers and Scientists Using MATLAB®

POWER METHODS

Find all eigenvalues and eigenvectors of the following matrix using the power, shifted inverse power, and shifted power methods:

$\mathbf{A}=\left[\begin{array}{cccc} 4 & -3 & 3 & -9 \\ -3 & 6 & -3 & 11 \\ 0 & 8 & -5 & 8 \\ 3 & -3 & 3 & -8 \end{array}\right], \quad \mathbf{x}_{1}=\left\{\begin{array}{l} 0 \\ 1 \\ 0 \\ 1 \end{array}\right\}$

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>> A = [4 −3 3 −9;−3 6 −3 11;0 8 −5 8;3 −3 3 −8];
>> x1 = [0;1;0;1]; % Initial vector
>> [e_val, e_vec] = PowerMethod(A, x1) % Default values for tol and kmax

e_val =

−5.0000 % Dominant eigenvalue

e_vec =

0.5000
−0.5000
−0.5000
0.5000

The smallest magnitude eigenvalue of $\mathbf{A}$ can be estimated by either applying the power method to $\mathbf{A}^{-1}$ (see Example 9.2) or by applying the shifted inverse power method to $\mathbf{A}$ with $\alpha=0$ .

>> [e_val, e_vec] = ShiftInvPower(A, 0, x1)

e_val =

1.0001 % Smallest magnitude eigenvalue

e_vec =

−0.8165
0.4084
0.0001
−0.4082

The largest and smallest eigenvalues of $\mathbf{A}$ are therefore -5 and 1, respectively. To see if the remaining two eigenvalues are between these two values, we set $\alpha$ to be the average, $\alpha=(-5+1) / 2=-2$ , and apply the shifted inverse power method. However, $\alpha=-2$ causes $\mathbf{A}-\alpha \mathbf{I}$ to be singular, meaning $\alpha$ is an eigenvalue of $\mathbf{A}$ . Since we also need the eigenvector associated with this eigenvalue, we set $\alpha$ to a value close to -2 and apply the shifted inverse power:

>> [e_val, e_vec] = ShiftInvPower(A, −1.5, x1) % alpha=−1.5

eigenval =

−2.0000 % Third eigenvalue

eigenvec =

0.5774
−0.5774
0.0000
0.5774

We find the fourth eigenvalue using the shifted power method. Knowing $\lambda=-2$ is an eigenvalue of $\mathbf{A}$ , we apply the power method to $\mathbf{A}+2 \mathbf{I}$ :

>> A1 = A + 2*eye(4,4);
>> [e_val, e_vec] = PowerMethod(A1, x1)

e_val =

5.0000 % Fourth eigenvalue = 5+(−2)=3

e_vec =

−0.0000
0.7071
0.7071
0.0000

By the reasoning behind the shifted power method, the fourth eigenvalue is $\lambda=-2+5=3$ . In summary, all four eigenvalues and their eigenvectors are

$\lambda_{1}=-5, \mathbf{v}_{1}=\left\{\begin{array}{l} 1 \\ -1 \\ -1 \\ 1 \end{array}\right\}, \quad \lambda_{2}=1, \mathbf{v}_{2}=\left\{\begin{array}{l} -2 \\ 1 \\ 0 \\ -1 \end{array}\right\}, \quad \lambda_{3}=-2, \mathbf{v}_{3}=\left\{\begin{array}{l} 1 \\ -1 \\ 0 \\ 1 \end{array}\right\}, \quad \lambda_{4}=3, \mathbf{v}_{4}=\left\{\begin{array}{l} 0 \\ 1 \\ 1 \\ 0 \end{array}\right\}$