Linear Algebra, Theory And Applications, 2012a

10.7. THE RATIONAL CANONICAL FORM 267
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 [9] and
there are more details given in this reference. Another good source which has many of the
same ideas is [14].
Here is a definition of the concept of a companion matrix.
Definition 10.7.1 Let
q (λ) =a 0 + a 1 λ + ···+ a n−1 λ n−1 + λ n
be a monic polynomial. The companion matrix of q (λ) , denoted as C (q (λ)) is the matrix
⎛
⎞
0 ··· 0 −a 0
1 0 −a 1
⎜
⎝
. .. . .. .
⎟
⎠
0 1 −a n−1
Proposition 10.7.2 Let q (λ) be a polynomial and let C (q (λ)) be its companion matrix.
Then q (C (q (λ))) = 0.
Proof: Write C instead of C (q (λ)) for short. Note that
Ce 1 = e 2 ,Ce 2 = e 3 , ··· ,Ce n−1 = e n
Thus
e k = C k−1 e 1 ,k=1, ··· ,n (10.11)
and so it follows {
e1 ,Ce 1 ,C 2 e 1 , ··· ,C n−1 }
e 1 (10.12)
are linearly independent. Hence these form a basis for F n . Now note that Ce n is given by
and from (10.11) this implies
Ce n = −a 0 e 1 − a 1 e 2 −···−a n−1 e n
C n e 1 = −a 0 e 1 − a 1 Ce 1 −···−a n−1 C n−1 e 1
and so
q (C) e 1 = 0.
Now since (10.12) 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) e 1 = k (C) q (C) e 1 = 0
which shows q (C) =0.
The following theorem is on the existence of the rational canonical form.
Theorem 10.7.3 Let A ∈L(V,V ) where V is a vector space with field of scalars F and
minimal polynomial
q∏
φ i (λ) mi
i=1