Linear Algebra, Theory And Applications, 2012a

248 LINEAR TRANSFORMATIONS CANONICAL FORMS
10.2 Direct Sums, Block Diagonal Matrices
Let V be a finite dimensional vector space with field of scalars F. Here I will make no
assumption on F. Also suppose A ∈L(V,V ) .
Recall Lemma 9.4.3 which gives the existence of the minimal polynomial for a linear
transformation A. This is the monic polynomial p which has smallest possible degree such
that p(A) =0. It is stated again for convenience.
Lemma 10.2.1 Let A ∈L(V,V ) where V is a finite dimensional vector space of dimension
n with field of scalars F. Then there exists a unique monic polynomial of the form
p (λ) =λ m + c m−1 λ m−1 + ···+ c 1 λ + c 0
such that p (A) =0and m is as small as possible for this to occur.
Now here is a useful lemma which will be used below.
Lemma 10.2.2 Let L ∈L(V,V ) where V is an n dimensional vector space. Then if L
is one to one, it follows that L is also onto. In fact, if {v 1 , ··· ,v n } is a basis, then so is
{Lv 1 , ··· ,Lv n }.
Proof: Let {v 1 , ··· ,v n } be a basis for V . Then I claim that {Lv 1 , ··· ,Lv n } is also a
basis for V . First of all, I show {Lv 1 , ··· ,Lv n } is linearly independent. Suppose
Then
and since L is one to one, it follows
n∑
c k Lv k =0.
k=1
( n
)
∑
L c k v k =0
k=1
n∑
c k v k =0
k=1
which implies each c k = 0. Therefore, {Lv 1 , ··· ,Lv n } is linearly independent. If there
exists w not in the span of these vectors, then by Lemma 8.2.10, {Lv 1 , ··· ,Lv n ,w} would
be independent and this contradicts the exchange theorem, Theorem 8.2.4 because it would
be a linearly independent set having more vectors than the spanning set {v 1 , ··· ,v n } .
Now it is time to consider the notion of a direct sum of subspaces. Recall you can
always assert the existence of a factorization of the minimal polynomial into a product of
irreducible polynomials. This fact will now be used to show how to obtain such a direct
sum of subspaces.
Definition 10.2.3 For A ∈L(V,V ) where dim (V )=n, suppose the minimal polynomial
is
q∏
p (λ) = (φ k (λ)) r k
k=1
where the polynomials φ k have coefficients in F and are irreducible. Now define the generalized
eigenspaces
V k ≡ ker ((φ k (A)) r k
)
Note that if one of these polynomials (φ k (λ)) r k
is a monic linear polynomial, then the generalized
eigenspace would be an eigenspace.