Here one has the minimal polynomial in the form ∏_{k=1}^{q}ϕ
(λ)
^{mk} where ϕ
(λ)
is an irreducible
monic polynomial. It is not necessarily the case that ϕ
(λ)
is a linear factor. Thus this case is
completely general and includes the situation where the field is arbitrary. In particular, it includes
the case where the field of scalars is, for example, the rational numbers. This may be partly
why it is called the rational canonical form. As you know, the rational numbers are
notorious for not having roots to polynomial equations which have integer or rational
coefficients.
This canonical form is due to Frobenius. I am following the presentation given in [10] and there
are more details given in this reference. Another good source which has additional results is
[15].
Here is a definition of the concept of a companion matrix.
e_{1} = 0. Now since 9.12 is a basis, every vector of F^{n} is of the form k
(C)
e_{1} for some
polynomial k
(λ)
. Therefore, if v ∈ F^{n},
q (C )v = q(C)k (C )e1 = k (C )q(C)e1 = 0
which shows q
(C )
= 0. ■
The following theorem is on the existence of the rational canonical form.
Theorem 9.7.3Let A ∈ℒ
(V,V )
where V is a vector space with field of scalars F andminimal polynomial∏_{i=1}^{q}ϕ_{i}
(λ)
^{mi}where each ϕ_{i}
(λ)
is irreducible and monic. LettingV_{k}≡ ker
(ϕk (λ)mk)
, it follows
V = V1 ⊕ ⋅⋅⋅⊕ Vq
where each V_{k}is A invariant. Letting B_{k}denote a basis for V_{k}and M^{k}the matrix of therestriction of A to V_{k}, it follows that the matrix of A with respect to the basis
{B1,⋅⋅⋅,Bq}
is theblock diagonal matrix of the form
( 1 )
| M 0 |
|( ... |) (9.13)
0 M q
(9.13)
If B_{k}is given as
{βv ,⋅⋅⋅,βv }
1 s
as described in Theorem 9.3.5where each β_{vj}is an A cyclic set ofvectors, then the matrix M^{k}is of the form
( )
C (ϕk(λ)r1) 0
M k = || .. || (9.14)
( . r )
0 C (ϕk(λ) s)
(9.14)
where the A cyclic sets of vectors may be arranged in order such that the positive integersr_{j}satisfy r_{1}≥
⋅⋅⋅
≥ r_{s}and C
rj
(ϕk(λ) )
is the companion matrix of the polynomialϕ_{k}
(λ)
^{rj}.
Proof: By Theorem 9.2.5 the matrix of A with respect to
{B ,⋅⋅⋅,B }
1 q
is of the form given in
9.13. Now by Theorem 9.3.5 the basis B_{k} may be chosen in the form
{β ,⋅⋅⋅,β }
v1 vs
where each
β_{vk} is an A cyclic set of vectors and also it can be assumed the lengths of these β_{vk} are decreasing.
Thus
Vk = span(βv1)⊕ ⋅⋅⋅⊕ span (βvs)
and it only remains to consider the matrix of A restricted to span
(βvk)
. Then you can apply
Theorem 9.2.5 to get the result in 9.14. Say
d−1
βvk = vk,Avk,⋅⋅⋅,A vk
where η
(A)
v_{k} = 0 and the degree of η
(λ)
is d, the smallest degree such that this is so,
η being a monic polynomial. Then η