[Octave-bug-tracker] [bug #38577] Eig returns non-unitary transformation matrix

[Rik]

Follow-up Comment #1, bug #38577 (project octave):

The eigenvalues in E are correct ([1, -1, lots of zeros]).  This can be
checked with sum (E(:)) which is always 0.

For the eigenvectors with zero eigenvalues one needs to solve Ax = 0*x.  The
zero vector for x solves this equation and is orthogonal to the other
eigenvectors which are all zeros except for a single '1' on the diagonal. 
Octave appears to place this as the very last eigenvector every time.  You can
check this with diag (U*U')(end-1).

So in this issue merely definitional?  Octave believes the zero vector is a
valid eigenvector because it meets the criteria above, but ordinary usage does
not include the zero vector as an eigenvector?



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