# Where does the Jordan canonical form show up in more advanced mathematics?

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?

• It is used to compute the exponential of a matrix, inthe context of linear differential equations. – Bernard Mar 10 '18 at 22:35
• "All I hear from the applied mathematicians is that the Jordan form is useless in practice". I think it is down to the applied mathematicians you have been talking to do justify their derogatory remarks. – Rob Arthan Mar 10 '18 at 22:45
• @RobArthan is it derogatory? In what sense? I'm interested in hearing your thoughts ... – D.Hutchinson Mar 10 '18 at 23:06
• The JNF is numerically unstable everywhere outside of diagonalizable matrices; I don't see how it could be of any direct numerical use. What I would expect to be useful are specific formulas for JNFs of specific types of matrices. – darij grinberg Mar 10 '18 at 23:07
• Did someone mention derogatory matrices? – Git Gud Mar 10 '18 at 23:12

In brief, the Jordan normal form and various of its siblings are ubiquitous in mathematics, whether labelled "pure" or "applied" or whatever. This is not to assert that numerical computation of literal Jordan forms is worthwhile, or stable, etc. In fact, as in my earlier comment, the very instability can play a very practical role in looking at families of linear systems that "go close" to the kind of degeneracy that non-trivial Jordan blocks depict.

For that matter, Jordan form is just a special case of the structure theorem for finitely-generated modules over principal ideal domains, such as $k[x]$ where $k$ is a field. Again, yes, in both a strong-topology sense and in a Zariski-topology sense, having non-trivial "Jordan blocks" is anomalous, but it can happen, and things "nearby" start to behave less stably.

Various non-normal compact operators on Hilbert spaces also can have non-trivial Jordan blocks. For example, the Volterra operator $Vf(x)=\int_0^x f(t)\,dt$ has this behavior.

The importance of rationality and algebraic-group-theoretic aspects of Jordan form in the theory of algebraic groups was already mentioned. This is a big deal, after all.

In general, "failure of semi-simplicity" is an awkward thing, and is to be proven not to happen (if that is the case). Failure is manifest in too-extensive non-trivial Jordan blocks. For example, "Galois representations" (that is, repns of Galois groups on various cohomologies of algebraic-geometric objects) "should be" semi-simple, etc., but this requires proof.

The case that a second-order ODE degenerates to have two closely-related solutions is an instance of a non-trivial Jordan block.

In complex analysis and algebraic geometry, representations of $\pi_1(X)$ as "monodromy groups" raise the potential issue of non-trivial Jordan blocks.

And so on. "Jordan block" is an essential descriptive notion, nearly everywhere. People who say it's "useless" are either just playing rhetorical games, or are pretty ignorant of serious mathematics.

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)

• It's also worth noting I think that without Jordan decomposition it would be impossible to define reductive groups, perhaps the most important class of linear algebraic groups. For the OP, reductive algebraic groups are of interest in part because their finite dimensional representations are completely reducible over $\mathbb{C}$. – leibnewtz Mar 11 '18 at 19:20
• Nice expository post about a fairly technical point in algebraic group theory! Let me add that its relevance doesn't start in algebraic group theory; the additive analogue of this decomposition ($x = x_s + x_n$ instead of $g = g_s g_n$) also figures in the algebraic proof of Weyl's theorem stating that any finite-dimensional module over a semisimple Lie algebra is completely reducible. – darij grinberg Mar 11 '18 at 23:32

Although the Jordan canonical form (JCF) may not be too useful for applied mathematicians, it is terribly important for theoretical mathematicians (see several consequences of the JNF), especially matrix theorists.

For instance, suppose that $f:\mathbb{C} \longrightarrow \mathbb{C}$ and $A$ is an $n$-by-$n$ matrix such that $A= S D S^{-1}$, with $\text{diag}(\lambda_1,\dots,\lambda_n)$. It is natural to define $f(A)$ as the $n$-by-$n$ matrix $Sf(D)S^{-1}$, with $f(D) = \text{diag}(f(\lambda_1),\dots,f(\lambda_n))$. What about the case when $A$ is defective? Is it possible to make sense of $f(A)$?

The answer is 'yes' and can be found in the book on matrix function theory by Higham. The JCF plays a central role in this book (and many treatises on matrix theory).

As you noted, the Jordan Canonical Form (JCF) is not very useful for applied math, but it is in fact important for pure math.

I'll make an example which doesn't touch very advanced topics. You know that JCF gives you a classification of endomorphisms of a vector space. Now, this consequently leads you to a classification of homographies, and gives you a way to study their fixed points, invariant subspaces etc.

For example, if you know that every endomorphism $f: \mathbb{C}^n \longrightarrow \mathbb{C}^n$ has a flag basis, you are already able to prove that every homography of a complex projective space has a fixed point.

More generally, it's easy to see that if the matrix which induces the homography has a JCF with $k$ blocks for the same eigenvalue, than the homography have a projective subspace of fixed points of dimension $k-1$.