proof of Jordan canonical form theorem
This theorem can be proved combining the cyclic decomposition theorem and the primary decomposition theorem.
By hypothesis, the characteristic polynomial
of T factorizes completely over F, and then so does the minimal polynomial
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
V=r⊕i=1ker((T-λiI)αi). |
Let Ti be the restriction of T to ker((T-λiI)αi). We apply now the cyclic decomposition theorem to every linear operator
(Ti-λiI):ker(T-λiI)αi→ker(T-λiI)αi. |
We know then that ker(T-λiI)αi has a basis Bi of the form Bi=B1,i⋃B2,i⋃…⋃Bdi,i such that each Bs,i is of the form
Bs,i={vs,i,(T-λi)vs,i,(T-λi)2vs,i,…,(T-λi)ks,ivs,i}. |
Let’s see that T in each of this “cyclic sub-basis” Bs,i is a Jordan block:
Simply notice the following fact about this polynomials
:
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
T(T-λiI)j(vs,i)=(T-λi)j+1(vs,i)+λi(T-λiI)j(vs,i). |
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
(λi00⋯001λi0⋯0001λi⋯00⋮⋮⋮⋱⋮⋮000⋯λi0000⋯1λi) |
So, taking the basis B=B1⋃B2⋃…⋃Br, 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 |