# For a repeated eigenvalue only one eigenvctor is being returned

3 views (last 30 days)
James on 17 Mar 2024
Moved: Bruno Luong on 14 Apr 2024
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.

Karl on 17 Mar 2024
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

Bruno Luong on 17 Mar 2024
Edited: Bruno Luong on 17 Mar 2024
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 on 17 Mar 2024
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 =

Bruno Luong on 14 Apr 2024
Moved: Bruno Luong on 14 Apr 2024
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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