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Zero eigenvalues or eigenvectors

  1. Feb 23, 2015 #1


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    I have a bit of problem with zero eigenvectors and zero eigenvalues. On one hand, there seems to be nothing in the definition that forbids them, and they even seem necessary to allow because an eigenvalue can serve as a measurement and zero can be a measurement, and if there is a zero eigenvalue then it will be a term in a diagonalized matrix, so that one has a zero eigenvector as well (a column vector of the diagonal matrix with the zero eigenvalue). So far, so good. But on the other hand, if the zero eigenvector is allowed, then every value in the field would be an eigenvalue, hence making it a bit trivial, no?
  2. jcsd
  3. Feb 23, 2015 #2


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    Eigenvectors are non-zero by definition.

    An eigenvalue of zero on the other hand is fine. If you have a zero column in your diagonal matrix, you have to chose a non-zero value for the entry which gets multiplied by the zero eigenvalue in order to get a proper eigenvector.
  4. Feb 23, 2015 #3
    I have never seen a zero eigenvector, but zero eigenvectors come up in the theory of vibrations often enough. In that context, it means that the system is positive semi-definite, rather than positive definite. In physical terms, it means that the stiffness matrix is singular and that rigid body motion is possible.
  5. Feb 23, 2015 #4


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    Thanks, kith and Dr. D.
    kith: that clears it up nicely, thanks.
    Dr. D. Good to know: except I presume you had a typo, in that you meant , instead of
    that "but zero eigenvalues come up in the theory of vibrations"
  6. Feb 24, 2015 #5


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    Note that, generally speaking, zero eigenvalues are nothing special. Given a matrix A with an eigenvector x,
    A x = \lambda x
    I can construct a new matrix ##B = A - \lambda I##, where ##I## is the identity matrix, such that
    B x = 0 x
    This is called eigenvalue shifting, and is used in numerical methods for eigenequations, in order to speed up convergence.
  7. Feb 24, 2015 #6


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    There is a slight ambiguity in the definition of "eigenvector". Most textbooks define "eigenvalue" for a linear operator, A, as a number, [itex]\lambda[/itex] such that there exist a non-zero vector v with [itex]Av= \lambda v[/itex] and then define an "eigenvector" corresponding to eigenvalue [itex]\lambda[/itex] as such a non-zero vector. But some define eigenvalue in that way and then define "eigenvector" as any vector, v, satisfying [itex]Av= \lambda v[/itex]. I prefer that- it allows one to say things like "any multiple of an eigenvector is an eigenvector" without having to say "except 0" and "the set of all eigenvectors corresponding to eigenvalue [itex]\lambda[/itex] form a vector space" without having to say "with the 0 vector added".
    Last edited by a moderator: Feb 24, 2015
  8. Feb 24, 2015 #7
    Nomadreid, yes, you are correct. I was thinking one thing and typing another.
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