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I am using a two different computational libraries to calculate the eigenvectors and eigenvalues of a symmetric matrix. The results show that the eigenvalues calculated with both libraries are exactly the same, however, the eigenvectors differ. Nevertheless, both seem to be correct since their eigenvectors are orthogonal and the factorization is also correct. How can that be possible? I asked the developers and although they seemed confused, they said that "probably" has to do with the directions of the vectors.

If it is possible, is there a kind of margin error between those two different range of eigenvectors so I could compare them and make sure they are in range (they are right)?

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    $\begingroup$ The eigenvalues must be the same, not the eigenvectors. For instance, $I_2$ has $(1,0)$ and $(0,1)$ as natural basis of eigenvectors. But $(1,1)$ and $(1,-1)$ is also a basis of eigenvectors for $I_2$. $\endgroup$
    – Julien
    Commented May 16, 2013 at 13:40
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    $\begingroup$ Have you ruled out that the dimension of the eigenspace is more than 1? $\endgroup$
    – abnry
    Commented May 16, 2013 at 13:40
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    $\begingroup$ @vadim123 A dimension one space too has many possible bases. $\endgroup$
    – Julien
    Commented May 16, 2013 at 13:42
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    $\begingroup$ If the developers themselves don't know, it is unlikely that anybody else will know how they designed their algorithm. $\endgroup$
    – Julien
    Commented May 16, 2013 at 13:47
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    $\begingroup$ @Manolete Probably because they're using slightly different algorithms to compute them... Without looking at the algorithms closely, it's impossible to say where exactly the difference lies, though. It's also rather irrelevant, the important thing to take away from this is that the eigenvectors are not uniquely determined by the matrix. $\endgroup$
    – fgp
    Commented May 16, 2013 at 13:56

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There is no promise of uniqueness for eigenvectors, (at least, not without a lot of constraints decided upon ahead of time.)

If $v$ is an eigenvector for $\lambda$, then so is $\alpha v$ for every $\alpha$ in your field $F$. It is not alarming to have different eigenvectors for a single eigenvalue.

It is also not alarming to have linearly independent eigenvectors with the same eigenvalue. In fact, since your matrix is symmetric, it is diagonalizable, and so an eigenvalue of multiplicity $n$ must have $n$ linearly independent eigenvectors for that eigenvalue.

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