# Eigen Vectors and Values using Matlab

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Diana on 25 Feb 2021
Answered: Bjorn Gustavsson on 25 Feb 2021
I'm using the following code to find the eigen vectors, now when I evaluate lambda manually I get dfferent numbers
A=[4 6 2;6 0 3;2 3 -1];
[lambda]=eig(A)
lambda=round(lambda,2);
L=length(lambda)
E1=rref(A - lambda(1)*eye(L),1e-14)
E2=rref(A - lambda(2)*eye(L),1e-14)
E3=rref(A - lambda(3)*eye(L),1e-14)
But I'm getting the trivial solution while I shouldn't

Bjorn Gustavsson on 25 Feb 2021
Why not calculating the eigenvectors at once?
[Ev,lambda]=eig(A)
Also, when rounding the eigenvalues, you're no longer guaranteed to work with the eigenvalues - then the trivial solution is most likely the only solution.
HTH

KSSV on 25 Feb 2021
A=[4 6 2;6 0 3;2 3 -1];
lambda=eig(A)
syms l
eqn = det(A-l*eye(3))==0 ;
solve(eqn,l) ;
l = double(vpasolve(eqn,l))
##### 2 CommentsShowHide 1 older comment
KSSV on 25 Feb 2021
[v,d,w]=eig(A)

R2020b

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