Using MATLAB to realize: power method, inverse power method (origin displacement)

Power law

Example

Use power method , Calculate the main eigenvalue of the following matrix and the corresponding eigenvector .

$A=\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right]$

Realization

The mathematical iteration formula of power method is ：

$take{v}^{(0)}̸=0,\alpha ̸=0,Make{u}^{(0)}=v^{(0)}$

$v^{(1)}=A{v}^{(0)}=A{u}^{(0)}$

$u^{(1)}=\frac{v^{(1)}}{max(v^{(1)})}=\frac{A{v}^{(0)}}{max(A{v}^{(0)})}$

$v^{(2)}=A{u}^{(1)}=\frac{A{v}^{(1)}}{max(A{v}^{(0)})}=\frac{A^2{v}^{(0)}}{max(A^2{v}^{(0)})}$

$u^{(2)}=\frac{v^{(2)}}{max(v^{(2)})}=\frac{A^2{v}^{(0)}}{max(A^2{v}^{(0)})}$

By analogy .

Use Matlab Realization ：

format long g;
v0 = [1;1;1];
u0 = [1;1;1];
A = [2,-1,0;-1,2,-1;0,-1,2];
v = A * u0;
u = v / norm(v, inf);
i = 0;
while norm(u - u0, inf) >= 1e-5
    u0 = u;
    v = A * u0;
    u = v / norm(v, inf);
    i ++;
end;
norm(v, inf)
i
u

To solve the need to ：

$i=8,u^{(9)}=(0.70711,-1,0.70711{)}^T,\lambda =max(v^{(9)})=3.41422$

Inverse power method

Example

The following matrices are known to have eigenvalues $\lambda$ Approximate value $p=4.3$ , Use the inverse power method of origin displacement , Find the corresponding eigenvector $u$ , And improve $\lambda$ .

$A=\left[\begin{array}{ccc}3& 0& -10\\ -1& 3& 4\\ 0& 1& -2\end{array}\right]$

Realization

Inverse power method , Is based on the power method , A method for finding the minimum eigenvalue is derived . Due to the nature of eigenvalues , It can be used to find the exact eigenvalue of an approximate value .

The mathematical iteration method is as follows ：

First , The matrix $A-pI$ Triangulate , It is convenient to find the solution of the equations later .

$A-pI=LU$

seek $v^{(k)}$ when , It is equivalent to finding two trigonometric equations .

$\{\begin{array}{l}L{y}^{(k)}=u^{(k-1)}\\ U{v}^{(k)}=y^{(k)}\end{array}$

And then there is ：

$u^{(k)}=\frac{v^{(k)}}{max(v^{(k)})}$

Use Matlab Realization ：

A = [3,0,-10;-1,3,4;0,1,-2];
I = eye(3,3);
p = 4.3;
u0 = [1;1;1];
v = inv(A - p * I) * u0;
u = v / norm(v, inf);
i = 0;
while norm(u - u0, inf) > 1e-5
    u0 = u;
    v = inv(A - p * I) * u0;
    u = v / norm(v, inf);
    i ++;
end;
i
u
x = p + 1 / norm(v, inf)

To solve the need to ：

$i=6,u^{(7)}=(-0.96606,1,0.15210{)}^T,\lambda =p+\frac{1}{max(v^{(k)})}=4.57447$