Linear Algebra, Theory And Applications, 2012a

270 LINEAR TRANSFORMATIONS CANONICAL FORMS
Then multiplying both sides by φ (A) pj−1 this yields
∑d−1
c i0 φ (A) pj−1 A i v j =0
i=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. Since this does not happen, it follows each c i0 =0.
Thus
∑d−1
p
∑ j−1
i=0 k=1
c ik φ (A) k A i v j =0
Now multiply both sides by φ (A) pj−2 and do a similar argument to assert that c i1 =0for
each i. Continuing this { way, all the c}
ik = 0 and this proves the claim.
∣
Thus the vectors α j 0 , ··· ∣ ∣∣
,αj d−1
are linearly independent and there are p j d = ∣β vj
)
of them. Therefore, they form a basis for span
(β vj
. Also note that if you list the
columns { in reverse } order starting from the bottom and going toward the top, the vectors
α j 0 , ··· ,αj d−1
yield Jordan blocks in the matrix of φ (A). Hence, considering all these
{
s
vectors α j 0 d−1} , ··· ,αj listed in the reverse order, the matrix of φ (A) with respect to
j=1
this basis of V is in Jordan canonical form. See Proposition 10.4.4 and Theorem 10.5.2 on
existence and uniqueness for { the Jordan}
form. This Jordan form is unique up to order of
the blocks. For a given j α j 0 , ··· ,αj d−1
yields d Jordan blocks of size p j for φ (A). The
sizeandnumberofJordanblocksofφ (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) p j
. Now here is the main
result.
Theorem 10.8.4 Let V be a vector space having 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 ∏ q
k=1 φ k (λ) m k
. Then recall from Corollary
10.2.4
V = V 1 ⊕···⊕V q
where V k =ker(φ k (A) m k
) . Also recall from Corollary 10.2.5 that the minimal polynomial
of the restriction of A to V k is φ k (λ) m k
. Now apply Lemma 10.8.3 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 10.8.5 Here is a matrix.
⎛
A = ⎝ 5 −2 1 ⎞
2 10 −2 ⎠
9 0 9
Find a similarity transformation which will produce the rational canonical form for A.