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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"?

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    $\begingroup$ You say you understand the usefulness of diagonalization. Some matrices cannot be diagonalized; but every (square) matrix over an algebraically closed field can be put in Jordan canonical form. A matrix can be diagonalized iff its Jordan canonical form is diagonal. $\endgroup$ – Charles Staats Jun 22 '13 at 5:24
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    $\begingroup$ As far as practical uses go, Jordan Canonical Form play an important role in first order differntial (and difference) equations on $n$ variables, where generalized eigenvectors/spaces make a common appearance. $\endgroup$ – Omnomnomnom Jun 22 '13 at 5:25
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    $\begingroup$ @CharlesStaats: It is often remarked that JCF is the next best thing to diagonalization. I'm (sort of) trying to ask why people say this. $\endgroup$ – nigel Jun 22 '13 at 5:29
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    $\begingroup$ As far as niceness goes, a love for Jordan Canonical Form would have to come from a love of nilpotent matrices, which are the motivating pieces behind all this anyway (remember that $A-\lambda I$ is nilpotent over the generalized eigenspace of $\lambda$). $\endgroup$ – Omnomnomnom Jun 22 '13 at 5:29
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    $\begingroup$ Consider "shift operators" - operators that send a vector $p_1$ to $p_2$ then $p_2$ to $p_3$, then $p_3$ to $p_4$, and so forth, then send the last vector in the group to zero (Sort of like a conveyor belt transporting vectors to their ultimate doom). It is simple to show that such shift operators are not diagonalizable. The Jordan normal form shows the converse: shifting is essentially the only reason why an operator is not diagonalizable. $\endgroup$ – Nick Alger Jun 26 '16 at 17:48
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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.

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  • $\begingroup$ But the problem is that the JCF computation comes at a high prize. While in theory these results are valuable, are there any practical applications you can share with us? $\endgroup$ – Domi May 28 '15 at 18:11
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    $\begingroup$ @Domi: The structure provided by JCF of a matrix $A$ can be used to deduce the structure of problems where $A$ is underlying, for example the ODE $x'(t) = Ax(t)$ which is solved by $x(t) = \exp(At)x_0$. Instead of calculating that matrix exponential, we can plug in knowledge of how the matrix exponential of a JCF looks like, as shown here for example. $\endgroup$ – Roland Oct 2 '17 at 15:11
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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.

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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!

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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.

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