where V is a vector space having field of scalars F, the above shows there
exists a rational canonical form for A. Could A have more than one rational canonical form?
Recall the definition of an A cyclic set. For convenience, here it is again.
Definition 9.8.1Letting x≠0 denote by β_{x}the vectors
{x,Ax, A2x,⋅⋅⋅,Am −1x}
where mis the smallest such that A^{m}x ∈span
( m−1 )
x,⋅⋅⋅,A x
.
The following proposition ties these A cyclic sets to polynomials. It is just a review of ideas
used above to prove existence.
Proposition 9.8.2Let x≠0 and consider
{ }
x,Ax,A2x,⋅⋅⋅,Am−1x
. Then this is an Acyclic set if and only if there exists a monic polynomial η
(λ)
such that η
(A )
x = 0 andamong all such polynomials ψ
(λ)
satisfying ψ
(A )
x = 0, η
(λ)
has the smallest degree.If V = ker
m
(ϕ (λ) )
where ϕ
(λ)
is monic and irreducible, then for some positive integerp ≤ m,η
(λ)
= ϕ
(λ)
^{p}.
The following is the main consideration for proving uniqueness. It will depend on what was
already shown for the Jordan canonical form. This will apply to the nilpotent matrix
ϕ
(A )
.
Lemma 9.8.3Let V be a vector space and A ∈ℒ
(V,V )
has minimal polynomial ϕ
(λ)
^{m}where ϕ
(λ)
is irreducible and has degree d. Let the basis for V consist of
{βv1,⋅⋅⋅,βvs}
where β_{vk}is A cyclic as described above and the rational canonical form for A is the matrixtaken with respect to this basis. Then letting
|βvk|
denote the number of vectors in β_{vk}, itfollows there is only one possible set of numbers
|βvk|
.
Proof: Say β_{vj} is associated with the polynomial ϕ
(λ)
^{pj}. Thus, as described above
|| ||
βvj
equals p_{j}d. Consider the following table which comes from the A cyclic set
In the above, α_{k}^{j} signifies the vectors below it in the k^{th} column. None of these vectors below the
top row are equal to 0 because the degree of ϕ
(λ)
^{pj−1}λ^{d−1} is dp_{j}− 1, which is less than p_{j}d and
the smallest degree of a nonzero polynomial sending v_{j} to 0 is p_{j}d. Also, each of these
vectors is in the span of β_{vj} and there are dp_{j} of them, just as there are dp_{j} vectors in
β_{vj}.
Claim:The vectors
{ j j }
α0,⋅⋅⋅,αd−1
are linearly independent.
Proof of claim: Suppose
d− 1pj−1
∑ ∑ c ϕ(A)kAiv = 0
i=0 k=0 ik j
Then multiplying both sides by ϕ
(A )
^{pj−1} this yields
d∑−1 pj− 1 i
ci0ϕ(A) A vj = 0
i=0
this is because if k ≥ 1, you have a typical term of the form
cikϕ(A )pj−1ϕ (A )kAivj = Aiϕ (A )k− 1cikϕ (A )pj vj = 0
Now if any of the c_{i0} is nonzero this would imply there exists a polynomial having degree smaller
than p_{j}d which sends v_{j} to 0. In fact, the polynomial would have degree d− 1 + p_{j}− 1. Since this
does not happen, it follows each c_{i0} = 0. Thus
d−∑ 1pj∑−1 k i
cikϕ(A) A vj = 0
i=0 k=1
Now multiply both sides by ϕ
(A)
^{pj−2} and do a similar argument to assert that c_{i1} = 0 for each i.
Continuing this way, all the c_{ik} = 0 and this proves the claim.
Thus the vectors
{ }
αj0,⋅⋅⋅,αjd−1
are linearly independent and there are p_{j}d =
| |
|βvj|
of them.
Therefore, they form a basis for span
(βv)
j
. Also note that if you list the columns in reverse order
starting from the bottom and going toward the top, the vectors
{ j j }
α 0,⋅⋅⋅,αd−1
yield Jordan
blocks in the matrix of ϕ
(A)
. Hence, considering all these vectors
{ }
αj,⋅⋅⋅,αj
0 d− 1
_{j=1}^{s} , each
listed in the reverse order, the matrix of ϕ
(A)
with respect to this basis of V is in Jordan
canonical form. See Proposition 9.4.4 and Theorem 9.5.2 on existence and uniqueness for
the Jordan form. This Jordan form is unique up to order of the blocks. For a given j
{ j j }
α0,⋅⋅⋅,αd−1
yields d Jordan blocks of size p_{j} for ϕ
(A)
. The size and number of Jordan
blocks of ϕ
(A )
depends only on ϕ
(A)
, hence only on A. Once A is determined, ϕ
(A)
is determined and hence the number and size of Jordan blocks is determined, so the
exponents p_{j} are determined and this shows the lengths of the β_{vj},p_{j}d are also determined.
■
Note that if the p_{j} are known, then so is the rational canonical form because it comes from
blocks which are companion matrices of the polynomials ϕ
(λ)
^{pj}. Now here is the main
result.
Theorem 9.8.4Let V be a vector spacehaving field of scalars F and let A ∈ ℒ
(V,V )
.Then the rational canonical form of A is unique up to order of the blocks.
Proof: Let the minimal polynomial of A be ∏_{k=1}^{q}ϕ_{k}
. Also recall from Corollary 9.2.4 that the minimal polynomial of the
restriction of A to V_{k} is ϕ_{k}
(λ )
^{mk}. Now apply Lemma 9.8.3 to A restricted to V_{
k}.
■
In the case where two n × n matrices M,N are similar, recall this is equivalent to the two
being matrices of the same linear transformation taken with respect to two different bases. Hence
each are similar to the same rational canonical form.
Example 9.8.5Here is a matrix.
( )
| 5 − 2 1 |
A = ( 2 10 − 2 )
9 0 9
Find a similarity transformation which will produce the rational canonical form forA.
The minimal polynomial is λ^{3}− 24λ^{2} + 180λ − 432. Why? This factors as
2
(λ − 6) (λ − 12)
Thus ℚ^{3} is the direct sum of ker
( 2)
(A − 6I)
and ker
(A − 12I)
. Consider the first of these. You
see easily that this is
( ) ( )
1 − 1
y|( 1 |) + z|( 0 |) ,y,z ∈ ℚ.
0 1
What about the length of A cyclic sets? It turns out it doesn’t matter much. You can start with
either of these and get a cycle of length 2. Lets pick the second one. This leads to the
cycle
where the last of the three is a linear combination of the first two. Take the first two as the first
two columns of S. To get the third, you need a cycle of length 1 corresponding to ker
(A − 12I)
.
This yields the eigenvector
( 1 − 2 3 )
^{T}. Thus
( )
− 1 − 4 1
S = |( 0 − 4 − 2 |)
1 0 3
Now using Proposition 8.3.10, the Rational canonical form for A should be
Find a basis such that if S is the matrix which has these vectors as columns S^{−1}AS is in rationalcanonical form assuming the field of scalars is ℚ.
First it is necessary to find the minimal polynomial. Of course you can find the characteristic
polynomial and then take away factors till you find the minimal polynomial. However, there is a
much better way which is described in the exercises. Leaving out this detail, the minimal
polynomial is
3 2
λ − 12λ + 64λ− 128
This polynomial factors as
( 2 )
(λ− 4) λ − 8λ + 32 ≡ ϕ1(λ)ϕ2(λ)
where the second factor is irreducible over ℚ. Consider ϕ_{2}
because the next vector involving A^{2} yields a vector which is in the span of the above two. You
check this by making the vectors the columns of a matrix and finding the row reduced echelon
form. Clearly this cycle does not span ker
(ϕ2(A ))
, so look for another cycle. Begin with a vector
which is not in the span of these two. The last one works well. Thus another A cycle
is
and you see this is in rational canonical form, the two 2 × 2 blocks being companion matrices for
the polynomial λ^{2}− 8λ + 32 and the 1 × 1 block being a companion matrix for λ − 4. Note that
you could have written this without finding a similarity transformation to produce it.
This follows from the above theory which gave the existence of the rational canonical
form.
Obviously there is a lot more which could be considered about rational canonical forms. Just
begin with a strange field and start investigating what can be said. One can also derive more
systematic methods for finding the rational canonical form. The advantage of this is you don’t
need to find the eigenvalues in order to compute the rational canonical form and it can often be
computed for this reason, unlike the Jordan form. The uniqueness of this rational canonical form
can be used to determine whether two matrices consisting of entries in some field are
similar.