for the system 
, 
is technically a solution. you should search for non-trivial solution. for this you should create null space of 
matrix. and because this phenomena is numerically very sensitive to matrix, by using default functions you almost always get nothing (empty null space). because the matrix become full-rank by small purturbation caused by approximating eigenvalues. (to be more precise 
isn't invertible so numerically condition number of it is very high, making it critically non stable for any purturbation)
, is technically a solution. you should search for non-trivial solution. for this you should create null space of matrix. and because this phenomena is numerically very sensitive to matrix, by using default functions you almost always get nothing (empty null space). because the matrix become full-rank by small purturbation caused by approximating eigenvalues. (to be more precise isn't invertible so numerically condition number of it is very high, making it critically non stable for any purturbation)A=[4,0,1,0;0,4,1,0;1,1,4,2;0,0,2,4];
lambdas = (roots(poly(A)));
T1 = A - lambdas(1) * eye(4); % A - lambda*I
null(T1)
the thing you can do is try numerically find null space of T1 or equally eigenvector of A. one method that is iterative itself is finding SVD decomposition of matrix. as MATLAB documentation mentioned even the null function use SVD algorithm. the rest is up to you for finding eigenvector, but for example :
[U,S,V] = svd(A);
[S , zeros(4,1) , diag(real(lambdas))]
as you can see the S is diagonal of our lambdas.
norm(T1 * V(:,1))
you can see columns of matrix V are very close to eigenvectors.
