Determine eigvector through eigvalue without using eig

Hello. I am writing a script that uses the power method to calculate the principal and two lowest (or however the oposite of principal is) eigvalues of a couple of matrices.

For the last part i want to plot the eigvectors of the corresponding eigvalues but i dont want to use eig and havent been able to successfully use find to determine the position of the eigvals and therefore find the corresponding eigvectors.

Long story short i am looking for a way, in general, to find the corresponding eigvector of a matrix by only knowing the matrix and one of its eigvalues.

P.S. I think the code is irrelevant but just in case

clear
clc
format short
f = @(x) 2+sin(pi*x)
n=100
N=[1:n];
for i=1:n;
    x(i)=i/n;
end
E=f(x);
K=-2*E;
L=E(1:n-1);
M=E(2:n);
A=diag(K)+diag(L,1)+diag(M,-1);
k=3000;
yo=ones(n,1);

 
for i=1:k
    dian=A*yo;
    yo=dian./max(abs(dian));
end
dian
disp('H megisth kata apolith timh idiotimh tou pinaka A einai:') 
max(abs(dian))
 max(abs(eig(A)))

B=(n^2)*A;
for i=1:k
    dianB=B*yo;
    yo=dianB./max(abs(dianB));
end
disp('H megisth kata apolith timh idiotimh tou pinaka B einai:') 
max(abs(dianB))
max(abs(eig(B)))


Binv=inv(B);
for i=1:k
    dianBinv=Binv*yo;
    yo=dianBinv./max(abs(dianBinv));
end
disp('H elaxisth kata apolith timh idiotimh tou pinaka B einai:') 
max(abs(dianBinv))
max(abs(eig(Binv)))

Answer 1

[Torsten]

By definition, the eigenvectors are the null space of A-lambda*I.

Thus you can get the eigenvectors by

null(A-lambda*eye(size(A,1)))

Best wishes

Torsten.

Answer 2

[Bruno Luong]

null(A - lambda * eye(size(A))