where is the error in code of Shifted Power Method?

Ash=[3 -1 0;-2 6 -2;0 -1 3];

v=zeros(3,10);

vT=zeros(10,3);

v(:,1)=[0;1;0];

vT(1,:)=[0;1;0];

l=6.9996;

Av=zeros(3,10);

AvT=zeros(10,3);

m=zeros(10,1);

error=zeros(10,1);

display('-------------------------------------------------------------------------------------------')

display(' k [v^k]^T [Ashifited v^k]^T mk+1 |error|')

display('-------------------------------------------------------------------------------------------')

for k=1:10

if error(k,1)<0.0005

Ash=Ash-l*eye(3);

Av(:,k)=Ash*v(:,k);

AvT(k,:)=(Av(:,k))';

m(k+1,1)=Av(1,k);

vT(k,:)=(v(:,k))';

error(k+1,1)=abs(m(k+1,1)-m(k,1));

v(:,k+1)=Av(:,k)/m(k+1,1);

l=m(k+1,1);

end

end

[Jeffrey Clark]

@sadiqah aldahan, you should not use MATLAB commands (e.g., error) as your variable names. See Variable Names and use exist to check. Also don't use display where disp should be used. But your real issue is that when k==3 v(:,k+1)=Av(:,k)/m(k+1,1) divides 0 by 0 which results in NaN being stored in v which in turn results in plenty of NaN in almost every variable.

Answer 1

[Amanjeet Pani]

Do not use MATLAB keywords (e.g., error) as your variable names.

To display on the matlab command window use disp().

Also when k==3, v(:,k+1)=Av(:,k)/m(k+1,1) divides 0 by 0 which results in NaN being stored in v which in turn results in plenty of NaN in almost every variable.