# Is there something deep in the fact that an endomorphism of a finite dimensional complex vector space has an eigenvector?

In my course of linear algebra I studied that if $V$ is a finite dimensional vector space on the complex field, then every endomorphism of $V$ has an eigenvector. The proof is simple: taken a polynomial $f$ that is null on the endomorphism $A$ (it exists because of the finite dimension of $V$), exploiting the algebraic closure of $C$ we are there.

I gave the following geometrical interpretation: such a space can't be fully twisted by any of its linear operators, we always have at least an invariant subspace of dimenson 1. This fact still appears a bit magical to me. Is there some deep or different reason for this in other parts of mathematics such as geometry or algebra? Are there other interesting geometrical consequences? Does it remain true if the space is infinite dimensional?

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As to the depth and significance of the existence of eigenvectors for finite-dimensional linear operators over $\mathbb{C}$, I think you have already identified the core algebraic reason: it is that the field $\mathbb{C}$ is algebraically closed. I think it is fair to call this a "deep fact" -- the task of giving a rigorous proof was the topic of Gauss's thesis work, and in fact by modern standards Gauss's (first) proof is still not completely rigorous. (Some feel this way, anyway -- there is not universal agreement here.) Whole books have been written on various proofs of this result.

Of course the depth here lies in the fact that the definition of the complex field is topological / analytic, ultimately relying on the completeness of $\mathbb{R}$. If you start with a field $K$, the following is not very deep:

Proposition: For a field $K$, the following are equivalent:
(i) Every linear endomorphism of a finite dimensional $K$-vector space has an eigenvector.
(ii) $K$ is algebraically closed.

To see (i) ⇒ (ii), use the fact that every monic polynomial of degree $n$ is the chracteristic and minimal polynomial of its companion matrix.

To answer your second question: no, there are linear operators on infinite dimensional $\mathbb{C}$-vector spaces without eigenvalues, even bounded linear operators. A fundamental example is the Hilbert space $L^2([0,1])$ of square integrable $\mathbb{C}$-valued functions on the unit interval. Then the multiplication operator $M: f \mapsto xf$ has norm $1$ but is easily seen to have no eigenvectors, since the equation $xf = \lambda f$ forces $f = 0$ almost everywhere.

In functional analysis, there is a suitable generalization of the notion of the set of eigen*values* of a linear operator on certain infinite dimensional spaces, namely the spectrum.

Addendum: inspired by damiano's comment, here is what is in some sense the simplest possible example of a linear operator on an infinite dimensional space without an eigenvector: Let $K$ be any field, and let $V$ be a vector space of countably infinite dimension with basis ${e_n}_{n=1}^{\infty}$. Then consider the shift operator $T$ on $V$, i.e., the unique linear operator such that for all $n \in \mathbb Z^+$, $T(e_n) = e_{n+1}$. $T$ is very simple and easy to visualize: it just happens never to "cycle back" on itself. Indeed, if $v = \sum_{n=1}^{\infty} a_n e_n \in V$ (with $a_n = 0$ for all but finitely many $n$), then assuming that $Tv = \lambda v$ gives

$\sum_{n=1}^{\infty} \lambda a_n e_n = \sum_{n=2}^{\infty} a_{n-1} e_n$.

In particular $\lambda a_1 = 0$. It is clear that the kernel of $T$ is $0$, so if $v \neq 0$, $\lambda \neq 0$ and thus $a_1 = 0$ and then the above equation implies that $a_n = 0$ for all $n$, i.e., $v = 0$. (If $V = K[x]$ is the space of polynomials with $K$-coefficients, then with respect to the natural basis $e_n = x^{n-1}$, multiplication by $x$ is the shift operator $T$.)

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An example requiring less background knowledge would be using the multiplication by $x$, but only on polynomials, rather than $L^2$ functions. – damiano Aug 22 '10 at 10:30
Very nice, thank you :) – user365 Aug 22 '10 at 18:04

### Topological argument

Suppose $A$ has no kernel. Then it induces an automorphism of the projective space $\mathbb P(V)$. Now the fact that A has an eigenvector follows from the (purely topological) Lefschetz formula (because $\mathbb P(V)$ has non-zero Euler characteristic and any linear transformation lies in the connected component of the identity).

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Compare #9 of Ilya Nikokoshev's answer to the MO question asking for proofs of FTA: mathoverflow.net/questions/10535/…. – Pete L. Clark Aug 22 '10 at 10:24
A small correction: the fact that the cohomology of projective space vanishes in odd degrees is not enough to guarantee that the Lefschetz number is non-zero. For instance, the antipodal map on a sphere has no fixed points, and even dimensional spheres have vanishing odd degree cohomology. – damiano Aug 22 '10 at 10:28
@damiano oh, you're right, of course (well, it's enough for linear operators, since they lie in the connected component of the identity map) – Grigory M Aug 22 '10 at 10:38
By "lies in the connected component of the identity," do you mean in the compact-open topology (i.e., "is homotopic to the identity")? – Charles Staats Aug 22 '10 at 13:12
@marcuz well, I could guess from the text of the question, actually; but maybe it will be useful for somebody else... – Grigory M Aug 22 '10 at 18:19