TITLE: Interpreting the Cayley-Hamilton theorem
QUESTION [27 upvotes]: The statement of the Cayley-Hamilton Theorem is fairly straight-forward.
I now know how to find characteristic polynomials from a given matrix (or at least a matrix with certain properties that I am unaware of!). 
I know that the eigenvalues of the matrix are roots of the polynomial. 
But what does having such a polynomial mean? Wikipedia says that the characteristic polynomial "...encodes several important properties of the matrix...", but once we have switched to "matrix form" of the equation, what can we conclude?   
In other words, what does the Theorem do for us, besides allowing us to say, "Hey, I know a matrix solution to this polynomial"?? Is there an abstraction of this in abstract algebra (rings, fields, etc.)?
Thanks for your time.

REPLY [12 votes]: Let $K$ be an algebraically closed field.
Let $V$ be a vector space over $K$ of dimension $n$.
Let $u\colon V \rightarrow V$ be a $K$-linear map.
Let $K[X]$ be the polynomial ring with one variable.
$V$ can be regarded as a $K[X]$-module as follows.
Let $f(X) \in K[X], x \in V$.
We define $f(X)x$ as $f(u)(x)$.
It is easy to see that $V$ is a finitely generated torsion $K[X]$-module.
Hence $V$ has a composition series:
$V = V_0 \supset V_1 \supset\cdots \supset V_{r-1} \supset V_r = 0$.
Since each $V_i/V_{i+1}$ is a simple $K[X]$-module, it is isomorphic to $K[X]/(X - \alpha_i)$, where $\alpha_i \in K$.
It is easy to see that $f(X) = \prod_i (X - \alpha_i)$ is the characteristic polynomial of $u$.
It is also easy to see that $f(X)V = 0$.
This is the Cayley-Hamilton theorem.
Now let's consider the following analogy of the above argument.
Let $G$ be a finitely generated torsion $\mathbb{Z}$-module, i.e. a finite abelian group.
$G$ has a composition series:
$G = G_0 \supset G_1 \supset\cdots \supset G_{r-1} \supset G_r = 0$.
Each $G_i/G_{i+1}$ is isomorphic to $\mathbb{Z}/p_i\mathbb{Z}$, where $p_i$ is a prime number.
It is easy to see that $|G|= \prod_i p_i$.
It is also easy to see that $|G|G = 0$.
Therefore, in our analogy, $|G|$ corresponds to a characteristic polynomial and $|G|G = 0$ corresponds to the Cayley-Hamilton theorem.