Motivation for Jordan Canonical Form

Solution 1:

The most generic answer: any time that we can reduce a problem over an incredibly general object (say, a matrix) to a problem in which we have more information at our fingertips (say, the same problem but over matrices that are in JCF), we make life easier - both in terms of proving theory and in terms of practical computations.

To be more specific to the situation at hand: the Jordan canonical form is sort of the next-best-thing to diagonalization. If the matrix is diagonalizable, then its JCF is diagonal; if it isn't, then what you get is at least block diagonal, and the blocks come in a predictable form.

Solution 2:

According to the Domi's comment, the good question is not "is the JCF really useful ?"; the good question is "is the JCF practically computable ?". We can consider two cases.

Case 1. One knows only a numerical approximation $\tilde{A}$ of our matrix $A\in M_n$. Then $\chi_{\tilde{A}}$, the characteristic polynomial of $\tilde{A}$ has $n$ distinct roots; note that if $A$ has in fact multiple roots, then the roots of $\chi_{\tilde{A}}$ are very close. But, no matter, we can "diagonalize" $\tilde{A}$ with a numerical calculation whose complexity is $\approx 40 n^3$. Clearly, the calculation may be unstable even if we use $QR$ method. Anyway, in such a case, the JCF is useless!

Case 2. The matrix $A$ is exactly known. For the sake of simplicity, assume that $n\geq 5$ and that $A$ is any matrix with entries in $\mathbb{Z}$. Then we decompose $\chi_A$ in irreducible and we know when $A$ has multiple eigenvalues and we know approximations of these eigenvalues. Thus we can deduce, a priori, the form of the JCF of $A$. In these conditions, we can obtain (with the same complexity as above) an approximation of the JCF of $A$.

PS. Obviously, we can obtain the EXACT Frobenius canonical form also in $O(n^3)$; depending on the issue, this form can be very useful.