# Eigenvector existence in complex space

1. Oct 22, 2007

### jostpuur

I'm reading a proof where there's a conclusion: "Since $zW\subset W$, there is an eigenvector $v\neq 0$ of z in W, $zv=\lambda v$." There W is a subspace of some vector space V, and z is a matrix, in fact a member of some solvable Lie algebra $\mathfrak{g}\subset\mathfrak{gl}(V)$. (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 $W\subset\mathbb{C}^n$ be a vector space, and A a matrix such that $AW\subset W$. Then there exists an eigenvector $v\in W$ of the matrix A, so that $Av=\lambda v$ for some $\lambda \in\mathbb{C}^n$.

2. Oct 22, 2007

### jostpuur

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.

3. Oct 22, 2007

### 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.

4. Oct 22, 2007

### jostpuur

The matrix

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

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

5. Oct 22, 2007

### HallsofIvy

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

6. Oct 22, 2007

### jostpuur

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

Last edited: Oct 22, 2007