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Let $V$ be a finite dimensional complex vector space. This question states:

Every endomorphism of $V$ has an eigenvector. The proof is simple: given a polynomial $f$ that is null on the endomorphism $A$ (it exists because of the finite dimension of $V$), exploiting the algebraic closure of $\mathbb{C}$ we are there.

Can someone help me elaborate on this?

  • Why does the finite dimension of $V$ imply that there exists such a polynomial exists?

  • If a polynomial $f$ exists, I am then not sure how to show there must be an eigenvector. I'm aware that if there is an eigenvector $v$, then its eigenvalue $\lambda$ will satisfy $f(\lambda)=0$, which certainly is possible by the algebraic closure of $\mathbb{C}$. I'm just unaware of how you can conclude that there is such a $v$ in the first place.

Robin
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1 Answers1

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Why does the finite dimension of $V$ imply that there exists such a polynomial exists?

Say $V$ has dimension $n$. The space of endomorphisms of $V$ has finite dimension $m$ (namely $m=n^2$ will do). Let $A$ be an endomorphism. Then $\{I, A, A^2, \dots A^{m}\}$ is linearly dependent since it consists of $m+1$ elements. So some linear combination of these is zero. That is a polynomial $f$ so that $f(A) = O$.

Why is there an eigenvector?

Suppose $f(A) = O$, and the degree of the polymomial $f$ is as small as possible, say $q$. WLOG, $f$ is monic. The complex polynomial $f(z)$ factors $(z-a_1)\dots(z-a_q)$ [possibly with repeated factors]. That is, $(A-a_1I)(A-a_2I)\dots(A-a_qI)=O$. Because $q$ is as small as possible, the smaller prduct $(A-a_2I)\dots(A-a_qI)$ is not $O$, so there is a vector $v$ with $u := (A-a_2I)\dots(A-a_qI)v \ne 0$. But then $(A-a_1I)u = 0$ and $a_1$ is an eigenvector for $A$.

GEdgar
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