Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvector existence in complex space

  1. Oct 22, 2007 #1
    I'm reading a proof where there's a conclusion: "Since [itex]zW\subset W[/itex], there is an eigenvector [itex]v\neq 0[/itex] of z in W, [itex]zv=\lambda v[/itex]." There W is a subspace of some vector space V, and z is a matrix, in fact a member of some solvable Lie algebra [itex]\mathfrak{g}\subset\mathfrak{gl}(V)[/itex]. (Could be irrelevant information, though.)

    This seemed a strange claim, because just because z maps W into W doesn't in general mean that the eigenvector exists like that. However, I noticed that I was unable to come up with a counter example if I allowed the W to be over the complex field. Is there a theorem that says something like this:

    Let [itex]W\subset\mathbb{C}^n[/itex] be a vector space, and A a matrix such that [itex]AW\subset W[/itex]. Then there exists an eigenvector [itex]v\in W[/itex] of the matrix A, so that [itex]Av=\lambda v[/itex] for some [itex]\lambda \in\mathbb{C}^n[/itex].
  2. jcsd
  3. Oct 22, 2007 #2
    hmhm... the equation

    \textrm{det}(A-\lambda\cdot 1)=0

    of course has complex solutions. But is this enough for the existence of the eigenvectors? The conclusion works in the other direction at least. If that determinant is non-zero, then eigenvectors don't exist.
  4. Oct 22, 2007 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Of course there is an eigenvector. That W is a subspace is irrelevant. W is a complex vector space and z restricts to an endomorphism of it, hence it has an e-vector. This is true if we replace C with any algebraically closed field. What is an e-vector? A root of the char poly. In particular t is an e-value of Z implies ker(Z-t) is not zere. Let v be any element of the kernel, hence (Z-t)v=0.
  5. Oct 22, 2007 #4
    The matrix

    A =
    0 & -1 \\
    1 & 0 \\

    gives linear mappings [itex]A:\mathbb{C}^2\to\mathbb{C}^2[/itex] and [itex]A:\mathbb{R}^2\to\mathbb{R}^2[/itex] so that [itex]A\mathbb{C}^2\subset\mathbb{C}^2[/itex] and [itex]A\mathbb{R}^2\subset\mathbb{R}^2[/itex]. I can see that in the complex case vectors (1,-i) and (1,i) are eigenvectors with eigenvalues [itex]\pm i[/itex], but it doesn't look like that the eigenvectors exist in the real case.
  6. Oct 22, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    To me, at least, the question is not whether eigenvectors exist but whether eigenvalues exist. If [itex]\lambda[/itex] is an eigenvalue of A, then there exist non-0 v such that Av= [itex]\lambda[/itex]v by definition of "eigenvalue".

    Yes, some matrices, over the real numbers, do not have real eigenvalues- and so, since we are talking "over the real numbers" do not have eigenvalues. Since the complex numbers are algebraically closed, the "eigenvalue equation" always has a solution and so a matrix, over the complex numbers, always has at least one (complex) eigenvalue and so eigenvectors.
  7. Oct 22, 2007 #6
    Okey, I missed the meaning of "algebraically closed" previously.
    Last edited: Oct 22, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Eigenvector existence in complex space
  1. Complex Eigenvectors (Replies: 1)