Motivation for Jordan Canonical Form I took linear algebra and understood the proof that linear operators on a vector space over an algebraically closed field have a Jordan Canonical Form. Why should I care about this theorem? I understand that it can be useful in doing some computations, but it seems that these computations are quite rare. 
Indeed, I am not puzzled by diagonalization or triangularization at all. They both have practical and theoretical uses, but even more than that, they just seem like nice things to have. Can someone explain why Jordan Canonical Form is a "nice thing to have"?
 A: 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.
A: The idea of the JCF is to get a a linear transformation as close as possible to acting like scalar multiplication. When you restrict a linear transformation to one of its eigenspaces, by definition it is acting by scalar multiplication. However, the eigenspaces of a linear transformation will add up to the whole space only if it is diagonalizable. So, instead we consider 'generalized eigenvectors,' i.e. vectors $v$ for which $(T-\lambda I)^kv = 0$ for some power $k$ and scalar $\lambda$, as a weaker form of acting by scalar multiplication. The JCF is then the matrix of a linear transformation with respect to a full basis of generalized eigenvectors. The diagonal elements of the JCF tell you the $\lambda$, and the size of the block tells you the $k$. The beauty of the JCF then is that, if the base field contains all the eigenvalues of the linear transformation, it always exists!
A: 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.
A: G. Strang gives good motivation==> From 26:43- in this link https://www.youtube.com/watch?v=z_zYQHmrh08. I hope this may help you.
