Eigenvector existence in complex space

[jostpuur]

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].

 

[jostpuur]

hmhm... the equation

[tex]
\textrm{det}(A-\lambda\cdot 1)=0
[/tex]

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.

 

[matt grime]

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.

 

[jostpuur]

The matrix

[tex]
A =
\left[\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array}\right]
[/tex]

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.

 

[HallsofIvy]

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.

 

[jostpuur]

Okey, I missed the meaning of "algebraically closed" previously.