TITLE: Deducing results in linear algebra from results in commutative algebra
QUESTION [14 upvotes]: Here are two examples of results which can be deduced from commutative algebra: 

Any $n\times n$ complex matrix is conjugate to a Jordan canonical matrix (can be proven using the structure theorem for modules over a PID, in this case $\mathbf{C}[T]$ - see for example these course notes).
Commuting matrices have a common eigenvector (this can be seen as a consequence of Hilbert's Nullstellensatz, according to Wikipedia).

My question is, does anyone know of other examples of results in linear algebra which can be deduced (non-trivially*) from results in commutative algebra?
More precisely: Which results about modules over fields can be optained via modules over more general commutative rings? (Thanks to Martin's comment below for suggesting this precision of the question).
.* by "non-trivially," I mean you have to go deeper than simply applying module theory to modules over a field.

REPLY [4 votes]: This answer doesn't contain anything new and it was already alluded via the JNF in the question, but let me elaborate this point of view:
Given a vector space $V$ over a field $K$, then an endomorphism $f : V \to V$ is the same as giving $V$ the structure of a left $K[T]$-module (which restricts to the given $K$-module structure). The multiplication with $T$ on the left is just $f$. This is just the most natural gadget when you think about polynomials $f^n + a_{n-1} f^{n-1} + \dotsc + a_0$ and their action on $V$. Many notions of linear algebra can be understood in more concise terms with the help of this module (and actually this can be used as a motivation for an introduction to modules in a linear algebra class):

A $f$-invariant subspace of $V$ is just a $K[T]$-submodule of $V$
That $V$ is $f$-cyclic just means that $V$ is cyclic as a $K[T]$-module, i.e. generated by one element, i.e. isomorphic to $K[T]/(p)$ for some $p$. We have that $V$ is finite-dimensional iff $p \neq 0$.
It immediately follows that $f$-invariant subspaces of $f$-cyclic spaces are also $f$-cyclic. Try to prove this without the language of modules.
That $V$ is $f$-indecomposable just means that it is indecomposable as a $K[T]$-module
More general and concisely: The arrow category of $\mathsf{Vect}_K$ is isomorphic to the category of $K[T]$-modules.
The minimal polynomial $p$ of $f$ is just the unique normed generator of the kernel of the ring homomorphism $K[T] \to \mathsf{End}_K(V)$ given by the module.
When you think about the endomorphism of $K[T]/(p)$ which multiplies with $T$ in terms of the canonical basis, you will arrive automatically at the companion matrix of $p$ and the relation with $f$-cyclic vector spaces. You don't have to "learn" this, it is just there.
The structure theorem for finitely generated modules over principal ideal domains implies the general normal form (which holds over every field $K$): Every endomorphism of a finite-dimensional vector space is similar to a direct sum of cyclic endomorphisms. In coordinate language: Every matrix is similar to a block matrix of companion matrices.
If $K$ is algebraically closed, you may reduce to polynomials of the form $(T-\lambda)^n$, but $K[T]/(T-\lambda)^n \cong K[T]/T^n$ has a very simple matrix, namely the Jordan block. Hence, the general normal form implies the Jordan normal form: Every matrix is similar to a block matrix of Jordan blocks.