The existence of an eigenvalue of a finite-dim matrix, in a vector space over an arbitrary field.

Question

Any square finite-dimensional nonzero matrix acting in a space $\mathbb V$ has at least one eigenvector. This is how this is proven for $\mathbb V$ an $N$-dim space over the field of complex numbers $\mathbb C$, with $N$ finite.

Consider a nonzero vector $ v$. Since the $ N+1$ vectors $$ v\,,\;{M}\,v\,,\;{M}^2 v\,,\;.\,.\,.\,,\;{M}^N v\;\, $$ cannot be linearly independent, there are complex $\;a_0\,,\;a_1\,,\;.\;.\;.\;,\;a_N\,$, not all zero, such that $$ \vec{0}= a_0\,v+a_1\,{M}\,v+\;.\,.\,.\;+\;a_N\,{M}^N v\,\;. $$ Build a polynomial $$ a_0+a_1\,z+\;.\,.\,.\;+\,a_N\,z^N=a\,(z- p_1)\;.\,.\,.\;(z- p_n) $$ with $a,\,p_j\in\mathbb{C}$ and $a\neq 0$.
Here $n=N$ and $a=a_N$ for $a_N\neq 0$, and $n< N$ for $a_N= 0$.
In this polynomial's matrix counterpart $$ \vec{0} = a_0\,v\;+\;a_1\,{M}\,v\;+\;.\,.\,.\;+\;a_N\,{M}^N v $$ $$ = (a_0\;{\textrm{Id}}\,+\,a_1{M}\,+\;.\,.\,.\;+\,a_N\,{M^N})\;v $$ $$ = a\,({M}- p_1 {\,\textrm{Id}})\;.\,.\,.\;({M} - p_n {\,\textrm{Id}})\;v\;\,, $$ at least one operator $ {M}- p_j{\,\textrm{Id}}\,$ is noninvertible. For a finite-dim $ {M}$, this makes $ p_j$ its eigenvalue. A vector on which $ {M}- p_j{\,\textrm{Id}}\,$ is noninvertible is an eigenvector. $\;\;$QED

QUESTIONS:
Is this proof valid in a finite-dim space $\mathbb V$ over an arbitrary closed field?
Is it valid in a finite-dim vector space over a ring (say, the ring $\mathbb H$ of quaternions)?

Answer 1

If you have a real matrix whose (complex) eigenvalues are all not real, then clearly it cannot have any real eigenvalue because these would also be eigenvalues over the complex numbers. Take for example $M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ whose complex eigenvalues are $\pm i$.

The proof you wrote uses that the field of complex numbers is algebraically closed, meaning that every non-constant polynomial has a root (and thus by induction can be written as a product of polynomials of degree $1$).

The proof works over every algebraically closed field, but whenever you consider a field for which there exists a polynomial $p$ without a root, then you can also find a matrix whose characteristic polynomial (and minimal polynomial) is $p$. One way to do this is the companion matrix of a polynomial. Such a matrix the has no eigenvalue since eigenvalues are roots of the characteristic polynomial.

Answer 2

The eigenvalue problem over $\mathbb{H}$ appears to be surprisingly difficult and delicate. I found Fuzhen Zhang's Quaternions and Matrices of Quaternions and Stefano De Leo, Giuseppe Scolarici, and Luigi Solombrino's Quaternionic Eigenvalue Problem which have relevant information.

Noncommutativity breaks much of usual linear algebra. It is still true that $\{ v, Mv, \dots \}$ is linearly dependent. This leads to an equation of the form $a(M) v = 0$ where

$$a(M) = \sum a_k M^k$$

is what we might call a "left polynomial" in $M$ (meaning the coefficients are on the left). It is not clear a priori that such a thing admits a factorization; it is known at least that $a(t) = \sum a_k t^k$ has a root over $\mathbb{H}$ (this is due to Niven), but noncommutativity spoils the rest of the argument, at least as far as I can tell; it's not clear that a root implies a factorization of $a(M)$ (I'm not even sure it implies a factorization of $a(t)$) since $M$ doesn't necessarily commute with the coefficients of $a$. Relatedly, the evaluation map $a \mapsto a(M)$ is no longer a homomorphism.

In fact over $\mathbb{H}$ we have to distinguish between left eigenvalues $Mv = \lambda v$ and right eigenvalues $Mv = v \lambda$, and the theory of these are quite different. Left eigenvalues apparently are both harder to understand and less useful. For example,

• multiplication by $M$ commutes with right scalar multiplication but not left, so left eigenvalues are not even invariant under conjugation, and
• relatedly, if $M$ has left eigenvalue $\lambda$ it does not follow that $M^2$ has left eigenvalue $\lambda^2$! We have $M^2 v = M \lambda v$ but this is not necessarily equal to $\lambda Mv = \lambda^2 v$ if $M$ and $\lambda$ don't commute. This is discussed in de Leo et al.

Left eigenvalues were apparently not proven to exist until 1985 (!) by Wood. The proof does not involve left polynomials at all and instead uses a "quaternionic winding number" calculation involving $\pi_3(GL_n(\mathbb{H}))$.

Right eigenvalues appear to have a nicer theory (contrary to the above two bullet points, they are invariant under conjugation and if $M$ has right eigenvalue $\lambda$ then $M^k$ has right eigenvalue $\lambda^k$) and there are $n$ particularly nice ones; see Theorem 5.4 in Fuzhen Zhang's paper.

Answer 3

The first part is valid, but there are higher degree irreducible polynomials over general fields, so we do not necessarily have an eigenvalue (= linear factor).

Making the necessary changes one ends up with ‘companion matrices’, and an ‘eigenspace’ (or rather the equivalent of the one-dimensional space spanned by a single eigen vector) is then a subspace which has no proper invariant subspaces (invariant=stable under the action of the original matrix).

Once one gets to this point, it is usually more convenient to change perspective and consider finite dimensional modules for the polynomial ring $k[x]$, which is a principal ideal domain (PID), and then use the general structure theory for modules over a PID. This is completely analogous to the theory of finite abelian groups (=finite modules for the PID $\mathbb Z$).