Linear Algebra and Analysis

9.7 Canonical Forms

Now let D = F

and let L ∈ℒ where V is a finite dimensional vector space over F. Define the multiplication of something in D with something in V as follows.

g(x)v ≡ g(L)v, L0 ≡ I

As mentioned above, V is a finitely generated torsion module and F

is a p.i.d. The non-invertible primes are polynomials irreducible over F. Letting be a basis for V, it follows that

V = Dz1 + ⋅⋅⋅+ Dzl

and so there exist irreducible polynomials over F,

_i=1ⁿ and corresponding positive integers k_i such that

( k1) ( kn) V = ker ϕ1 (L) ⊕ ⋅⋅⋅⊕ ker ϕn (L )

This is just Theorem 6.1.10 obtained as a special case. Recall that the entire theory of canonical forms is based on this result. This follows because

{ } Mϕi = v ∈ V : ϕi(x)ki v ≡ ϕi(L)ki v = 0 .

Now continue letting D = F

,L ∈ℒ for V a finite dimensional vector space and gv ≡ gv as above. Consider M_ϕi ≡ ker which is a sub module of V . Then by Theorem 9.5.9, this is the direct sum of cyclic sub modules.

M ϕi = Dv1 ⊕ ⋅⋅⋅⊕ Dvs (*)

(*)

where s = rank

.

At this point, note that ann

= and so ann = where l_j ≤ k_i. If d_i is the degree of ϕ_i, this implies that for v_j being one of the v in ∗,

2 l d−1 1,Lvj,L vj,⋅⋅⋅,L ji vj

must be a linearly independent set since otherwise, ann

≠ because a polynomial of smaller degree than the degree of ϕ_i^l_j will annihilate v_j. Also,

( ) Dvj ⊆ span vj,Lvj,L2vj,⋅⋅⋅,Lljdi− 1vj

by division. Vectors in Dv_j are of the form g

v_j = v_j = ρv_j where the degree of ρ is less than l_jd_i. Thus a basis for Dv_j is and the span of these vectors equals Dv_j. Now you see why the term “cyclic” is appropriate for the submodule Dv. This shows the following theorem.

The last claim of this theorem will mean that the various canonical forms are uniquely determined.

It is clear that the span of β_vj^l_jd_i−1 is invariant with respect to L because, as discussed above, this span is Dv_j where D = F

. Also recall that ann = where l_j ≤ k_i. Let

l ϕi(x)j = xljdi + an− 1xljdi−1 + ⋅⋅⋅+ a1x+ a0

Recall that the minimum polynomial has leading coefficient equal to 1. Of course this makes no difference in the above presentation because a_n is invertible and so the ideals and above direct sum are the same regardless of whether this leading coefficient equals 1, but it is convenient to let this  happen since otherwise, the blocks for the rational canonical form will not be standard. Then what is the matrix of L restricted to Dv_j?

( ljdi−1 ljdi ) ( ljdi−2 ljdi− 1 ) Lvj ⋅⋅⋅ L vj L vj = vj ⋅⋅⋅ L vj L vj M

where M is the desired matrix. Now ann

= and so

( ) Lljdivj = (− 1)an− 1Lljdi− 1vj +⋅⋅⋅+ a1Lvj + a0vj

Thus the matrix M must be of the form

( ) 0 − a0 || 1 − a1 || || .. .. || (9.2) ( . . ) 0 1 − an−1

(9.2)

It follows that the matrix of L with respect to the basis obtained as above will be a block diagonal with blocks like the above. This is the rational canonical form.

Of course, those blocks corresponding to ker

can be arranged in any order by just listing the β_v1^l₁d_i−1,,β_{v si}^l_sid_i−1 in various orders. If we want the blocks to be larger in the top left and get smaller towards the lower right, we just re-number it to have l_i be a decreasing sequence.  Note that this is the same as saying that ann ⊆ ann ⊆⊆ ann. If we want the blocks to be increasing in size from the upper left corner to the lower right, this corresponds to re-numbering such that ann ⊇ ann ⊇⊇ ann. This second one involves letting l₁ ≤ l₂ ≤≤ l_si.

What about uniqueness of the rational canonical form given an order of the spaces ker

and under the convention that the blocks associated with ker should be increasing or degreasing in size from upper left toward lower right? In other words, suppose you have

( ki) ker ϕi(L) = Dv1 ⊕ ⋅⋅⋅⊕ Dvs = Dw1 ⊕ ⋅⋅⋅⊕ Dwt

such that

ann (Dv1) ⊆ ann (Dv2) ⊆ ⋅⋅⋅ ⊆ ann (Dvs )

and

ann(Dw1 ) ⊆ ann (Dw2) ⊆ ⋅⋅⋅ ⊆ ann (Dwt)

will it happen that s = t and the that the blocks associated with corresponding Dv_i and Dw_i are the same size? In other words, if ann

= and ann = is m_j = l_j. If this is so, then this proves uniqueness of the rational canonical form up to order of the blocks. However, this was proved above in the discussion on uniqueness, Theorem 9.6.1.

In the case that the minimum polynomial splits the following is also obtained.

Proof: The proof is essentially the same.

( ) ker (L− μ )ki = Dv ⊕ ⋅⋅⋅⊕ Dv i 1 si

ann

for v_j ∈ ker is a principal ideal where ν∕^k_i and so it is of the form ^l_j where 0 ≤ l_j ≤ k_i. Then as before,

2 lj−1 vj,(L− μiI)vj,(L − μiI) vj,⋅⋅⋅,(L − μiI) vj (*)

(*)

must be independent because if not, there is a polynomial g

of degree less than l_j such that g = 0 and so cannot really be ann. It is also true that the above list of vectors must span Dv_j because if f ∈ D, then f = ^l_jm + r where r has degree less than l_j. Thus fv_j = rv_j and clearly r can be written in the form with the same degree. (Just take Taylor expansion of r formally.) Thus ∗ is indeed a basis of Dv_j for v_j ∈ ker. ■

This gives the Jordan form right away. In this case, we know that

^l_jv_j = 0 and so the matrix of the transformation L − μ_iI with respect to this basis on Dv_j obtained in the usual way.

( lj−1 ) 0 (L − μiI) vj ⋅⋅⋅ (L− μiI)vj =

( ) 0 1 0 ( )|| .. || (L − μiI)lj−1vj (L − μiI)lj−2vj ⋅⋅⋅ vj || 0 . || |( ... 1 |) 0 0

a Jordan block for the nilpotent matrix

. Thus, with respect to this basis, the block associated with L and β_vj^l_j−1 is just

( ) | μi 1 0 | || μi ... || || .. || ( . 1 ) 0 μi

This has proved the existence of the Jordan form. You have a string of blocks like the above for ker

^k_i. Of course, these can be arranged so that the size of the blocks is decreasing from upper left to lower right. As with the rational canonical form, once it is decided to have the blocks be decreasing (increasing) in size from upper left to lower right, the Jordan form is unique.