My bounty for this question expires soon 🙂
Edit: in regards to the bounty offered, what current research trends use the Jordan canonical form?
If one takes a second course in Linear Algebra — or a graduate level Linear Algebra course — one typically learns about non-diagonalizable operators and the Jordan canonical form.
However, where does the Jordan canonical form show up again in later, more advanced mathematics? All I hear from applied mathematicians is that the Jordan canonical form is useless in practice (in academic research). If it's not useful in applied mathematics, is it an important tool in pure mathematics? If so, in which areas of pure mathematics?
Best Answer
Jordan canonical form is extremely important in the structure theory of linear algebraic groups.
Algebraic groups are of interest in many areas of pure mathematics, showing up in representation theory, algebraic geometry, number theory, and differential geometry (an algebraic group is the algebro-geometric analogue of a Lie group, and in fact all classical Lie groups can be regarded as algebraic groups), .
To make life easy, suppose $k$ is an algebraically closed field, and consider the group $\operatorname{GL}_n$ of $n$ by $n$ invertible matrices with entries in $k$. Then $\operatorname{GL}_n$ has a topology, called the Zariski topology, whose basic open sets are of the form
$$\{ (x_{ij}) \in \operatorname{GL}_n : \frac{f(x_{ij})}{\det(x_{ij})^m} \neq 0 \}$$
where $f$ is a polynomial in $n^2$ variables with coefficients in $k$, and $m$ is a nonnegative integer. A linear algebraic group is a closed subgroup of some $\operatorname{GL}_n$.
Examples: $\operatorname{GL}_1$ is just the group of nonzero elements of $k$. $\operatorname{SL}_n$ is the group of determinant one matrices. $\operatorname{SO}_n$ is the group of determinant one matrices whose inverse is their transpose.
Let $G \subseteq \operatorname{GL}_n$ and $H \subseteq \operatorname{GL}_m$ be linear algebraic groups. A morphism of algebraic groups is a group homomorphism $f: G \rightarrow H$ which is also a morphism of varieties. That is, for each $(x_{ij}) = x \in G$, $f(x)$ returns you an $m$ by $m$ matrix, and the entries of this matrix must be functions of $x$ of the form $\frac{f(x_{ij})}{\det(x_{ij})^p}$ for $f$ a polynomial and $p \geq 0$.
Here is one way to state Jordan canonical form.
Theorem: Let $A$ be an $n$ by $n$ matrix. There are unique matrices $A_s$ and $A_n$, such that $A_s$ is diagonalizable, $A_n$ is nilpotent, $A_s A_n = A_nA_s$, and $A = A_s + A_n$.
And here is the "multiplicative" version of Jordan canonical form, which follows directly from the usual version.
Theorem: Let $g \in \operatorname{GL}_n$. There are unique matrices $g_s, g_u$ such that $g_s$ is diagonalizable, $g_u$ is unipotent (i.e. $I - g_u$ is nilpotent), and $g = g_sg_u = g_u g_s$.
Now here is a remarkable theorem for linear algebraic groups, which is false for arbitrary subgroups of $\operatorname{GL}_n$:
Remarkable theorem: Let $G \subseteq \operatorname{GL}_n$ be a linear algebraic group. Let $g \in G$. Then $g_s$ and $g_u$ are in $G$.
Reference: T.A. Springer, Linear Algebraic Groups, Theorem 2.4.8
There is no immediate reason why $G$ should contains the matrices $g_s$ and $g_u$. Even though $g \in G$, the matrices $g_s$ and $g_u$ are a priori just some $n$ by $n$ invertible matrices which multiply to $g$. Yet $G$ must contain them.
Moreover, the notion of an element of a linear algebraic group being diagonalizable, or unipotent, exists independently of a particular realization of $G$ as a group of matrices! That is, if $H \subseteq \operatorname{GL}_m$ is a linear algebraic group, and $\phi: G \rightarrow H$ an isomorphism of algebraic groups, then for each $g \in G$, we have $\phi(g_s) = \phi(g)_s$ and $\phi(g_u) = \phi(g)_u$. Thus $\phi(g_s)\phi(g_u)$ is the Jordan decomposition of $\phi(g)$.
For an example of where Jordan canonical form is used in the structure theory, let $G$ be a linear algebraic group which is connected and solvable (for example, upper triangular matrices with nonzero diagonal elements). Let $G_u$ be the set of unipotent elements of $G$ (as mentioned in the previous paragraph, the notion of an element being unipotent does not depend on the specific realization of $G$ as a group of matrices). There exist maximal closed, connected subgroups of $G$ consisting of diagonalizable elements, called maximal tori, and if $T$ and $S$ are two maximal tori of $G$, then there exists a $g \in G$ such that $gTg^{-1} = S$. Then $G_u$ is a closed, connected, normal subgroup of $G$, and if $T$ is any maximal torus of $G$, then $G$ is the semidirect product of $G_u$ and $T$. (Reference: Springer 6.3.5)