For a repeated eigenvalue only one eigenvctor is being returned

for my matrix

A = [1 1 4 0; 1 1 1 -1; 0 0 3 1; 0 0 3 1], it is showing there is a repeated eigenvalue of 0, which is correct however the eigenvectors from when i do [V,D] = eig(A) for the 0's are both [-1; 1; 0; 0] when i have calculated there is a second one of [4/3; 0; -1/3; 1] just confused to why it is not outputting the eigenvectors.

Answer 1

[Karl]

Another approach, which gives the additional eigenvector that you calculated, is to obtain the eigenvector(s), x, for eigenvalue v as solutions of (A-v*I)*x = 0:

format rational
A = [1 1 4 0; 1 1 1 -1; 0 0 3 1; 0 0 3 1];
display_eigenvectors(A)

eigenvalue: 0.000000
 eigenvector(s):
      -1              4/3     
       1              0       
       0             -1/3     
       0              1       


eigenvalue: 2.000000
 eigenvector(s):
       1       
       1       
       0       
       0       


eigenvalue: 4.000000
 eigenvector(s):
       3/2     
       1/2     
       1       
       1       

function display_eigenvectors(A)
%DISPLAY_EIGENVECTORS Display eigenvectors for square matrix A.
%   The eigenvector(s), x, for each eigenvalue, v, are obtained as
%   a rational orthnormal basis of the null space of (A-v*I), where
%   I is the unit matrix with the same size as A.  The eigenvectors
%   are then solutions of (A-v*I)*x = 0.
I = eye(size(A));
for v=unique(eig(A))'
    fprintf(1,'\neigenvalue: %f\n eigenvector(s):\n',v)
    disp(null(A-v*I,'rational'))
end
end

Answer 2

[Bruno Luong]

Indeed in case of eigenvalue with multiplicity > 1; the problemie is numerical challenging and MATLAB might fail to find the correct eigen vectors as with your case.

[Bruno Luong]

The issue is that MATLAB numerical error will make matrix reduces to Jordan form and think that 0 has incorrectly 1-dimentional eigenspace and not 2 due to tiny numerical error.

Symbolic eig would work since there is no roundoff error

A = [1 1 4 0; 1 1 1 -1; 0 0 3 1; 0 0 3 1];
[V,D] = eig(sym(A))

V = 

D = 

Answer 3

[Bruno Luong]

UPDATE: From the discussion here using EIG with 2 arguments can do the trick and overcome the issue and return an independent eigen vector associate with 0

A = [1 1 4 0; 1 1 1 -1; 0 0 3 1; 0 0 3 1];
[V,D] = eig(A,eye(size(A)));
V4 = V(:,4)

V4 = 4x1

0.6667 0.6667 -0.3333 1.0000

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A*V4

ans = 4x1

0 0 0 0

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