## 403/503 – Eigenvectors for operators on real odd dimensional spaces

The goal of this note is to give a proof of the following result:

Theorem 1 Let ${V}$ be an odd dimensional vector space over ${{\mathbb R},}$ and let ${T:V\rightarrow V}$ be linear. Then ${T}$ admits an eigenvector.

The proof that follows is in the spirit of Axler’s textbook, so it avoids the use of determinants. However, I feel it is easier than the argument in the book, and it has the additional advantage of not depending on the fundamental theorem of algebra. In fact, the motivation for finding this argument was to avoid the use of the fundamental theorem.

The proof we present of Theorem 1 can be seen as an elaboration of the argument in the case when ${{\rm dim}(V)=3,}$ that we discussed in lecture. It was found by David Milovich in Facebook.

Proof: The argument is by induction on ${n={\rm dim}(V),}$ with the case ${n=1}$ being obvious.

Assume then that ${n}$ is odd, that the result holds for all positive odd dimensions smaller than ${n,}$ and that ${T:V\rightarrow V}$ is linear. We may also assume that the result is false for ${n}$, and argue by contradiction. It follows that ${T-\lambda I}$ is invertible for all ${\lambda\in{\mathbb R}}$ (where ${I}$ is the identity operator), and that if ${W\le V}$ is a proper ${T}$-invariant subspace, then ${{\rm dim}(W)}$ is even.

Lemma 2 Any ${v\in V}$ is contained in a proper ${T}$-invariant subspace.

Proof: This is clear if ${v=0.}$ Otherwise, recall that since ${v,Tv,\dots,T^n v}$ are linearly dependent, there is a polynomial ${p\in{\mathbb R}[x]}$ such that ${p(T)v=0}$ and ${p}$ is non-constant and of degree ${k\le n.}$ If ${k we are done, because then ${{\rm span}\{v,Tv,\dots,T^{k-1}v\}}$ is ${T}$-invariant and of dimension at most ${k.}$ If ${k=n,}$ we use that ${p}$ has a real root, to factor ${p(T)=(T-\lambda I)q(T)}$ for some ${\lambda\in{\mathbb R}}$ and ${q\in{\mathbb R}[x]}$ of degree ${n-1.}$ Since ${T-\lambda I}$ is invertible, we have that ${q(T)v=0}$ and, since ${{\rm deg}(q) we are done. $\Box$

Let ${W}$ be a maximal proper ${T}$-invariant subspace. let ${v\notin W,}$ and let ${U}$ be a proper ${T}$-invariant subspace with ${v\in U.}$ Note that ${W+U}$ is also ${T}$-invariant, and strictly larger than ${W,}$ since it contains ${v.}$ By maximality of ${W,}$ we must have ${W+U=V.}$ Since ${W}$ and ${U}$ have even dimension, it follows that ${{\rm dim}(W\cap U)}$ is odd. Since ${W\cap U}$ is ${T}$-invariant and proper, the inductive hypothesis gives us a contradiction, and we are done. $\Box$

Now, combining this proof with the results in The fundamental theorem of algebra and linear algebra by Harm Derksen, American Mathematical Monthly, 110 (7) (2003), 620-623 (also available through JSTOR) or in the preprint The fundamental theorem of algebra via linear algebra by Keith Conrad, we have a proof of the existence of eigenvectors for operators on finite dimensional complex vector spaces that does not require the use of determinants and does not use the fundamental theorem of algebra. By means of a well-known trick (the observation before the statement of Theorem 2 in Conrad’s paper, or Corollary 8 in Derksen’s), we actually obtain (once determinants are introduced) the fundamental theorem of algebra as a corollary of this purely linear algebraic result. Of course, at the heart of this proof are the two facts that any odd degree polynomial with real coefficients has a real root (used in the proof of Theorem 1), and that any complex number admits a complex square root, which are needed in all proofs of the fundamental theorem of algebra.

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