I am trying to code the inverse power method but the outcome of the eigenvector was half of the correct answer. Is there a way to fix it?

clc,clear
format long
A = input('Please enter square matrix A: ');
u = input('Please enter initial guess vector u: ');
s = input('Please enter max number of steps s: ');
[row,column] = size(A);
I = 1;
k = 0;
while I <= column
    if u(I) == 0
        I = I+1;
    elseif u(I) ~= 0
        y = sign(u(I))*(u/norm(u));
        I = 1;
        break
    end
end
while k <= s-1
    t = rref([A,y]);
    z = t(:,row+1);
    while I <= column
        if y(I)==0 & z(I)==0
            I = I+1;
        elseif y(I)~=0 & z(I)== 0
            I = I+1;
        elseif y(I)==0 & z(I)~=0
            I = I+1;
        elseif y(I)~= 0 & z(I)~=0
            p = y(I)/z(I);
            sig = sign(z(I));
            y = sig*(z/norm(z));
            k = k+1;
            break
        end
    end
end
fprintf('The approximation for the eigenvalue of A with smallest absolute value is: %s\n',p)
fprintf('The approximation for the eigenvector of A with smallest absolute value is:\n')
fprintf('                                 %d\n',[y])

Here is the example for the code:

%The matrix of A is:
           [4 1 0; 1 4 1; 0 1 4]
%The initial vector is:
           [0.05346163504453; 0.52970019333516; 0.67114938407724]
%The number of steps is:
            1000
%The supposed answer is:
            [1; -sqrt(2); 1]
%The answer I got from the code is:
            [0.5; -sqrt(2)/2; 0.5]

Answer 1

[Walter Roberson]

eigenvectors are direction but not magnitude. If v is an eigenvector of a matrix then so is m*v for all nonzero scalar m. Therefore you can choose to take the result you get and normalize it by dividing through by the first nonzero entry so that the first nonzero entry becomes 1.

[Kien Nguyen]

Oh thank you for pointing that out. I just forgot that the eigenvector can be scaled with the nonzero scalar m.