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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

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    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


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    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
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