proof of Jordan canonical form theorem

This theorem can be proved combining the cyclic decomposition theorem and the primary decomposition theoremPlanetmathPlanetmath. By hypothesis, the characteristic polynomialMathworldPlanetmathPlanetmath of T factorizes completely over F, and then so does the minimal polynomialPlanetmathPlanetmath of T (or its annihilator polynomial). This is because the minimal polynomial of T has exactly the same factors on F[X] as the characteristic polynomial of T. Let’s suppose then that the minimal polynomial of T factorizes as mT=(X-λ1)α1(X-λr)αr. We know, by the primary decomposition theorem, that


Let Ti be the restriction of T to ker((T-λiI)αi). We apply now the cyclic decomposition theorem to every linear operator


We know then that ker(T-λiI)αi has a basis Bi of the form Bi=B1,iB2,iBdi,i such that each Bs,i is of the form


Let’s see that T in each of this “cyclic sub-basis” Bs,i is a Jordan blockMathworldPlanetmath: Simply notice the following fact about this polynomialsPlanetmathPlanetmath:

X(X-λi)j = (X-λi)j+1+X(X-λi)j-(X-λi)j+1
= (X-λi)j+1+(X-X+λi)(X-λi)j
= (X-λi)j+1+λi(X-λi)j

and then


So, if we also notice that (T-λiI)ks,i+1(vs,i)=0, we have that T in this sub-basis is the Jordan block


So, taking the basis B=B1B2Br, we have that T in this basis has a Jordan form.

This form is unique (except for the order of the blocks) due to the uniqueness of the cyclic decomposition.

Title proof of Jordan canonical form theorem
Canonical name ProofOfJordanCanonicalFormTheorem
Date of creation 2013-03-22 14:15:36
Last modified on 2013-03-22 14:15:36
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 11
Author CWoo (3771)
Entry type Proof
Classification msc 15A18