Linear Algebra, Theory And Applications, 2012a

246 LINEAR TRANSFORMATIONS CANONICAL FORMS
It follows span (x 1 , ··· ,x n ,y 1 , ··· ,y m ) ⊇ ker (BA) and so by the first part, (See the picture.)
dim (ker (BA)) ≤ n + m ≤ dim (ker (A)) + dim (ker (B))
Of course this result holds for any finite product of linear transformations by induction.
One way this is quite useful is in the case where you have a finite product of linear
transformations ∏ l
i=1 L i all in L (V,V ) . Then
( )
l∏
l∑
dim ker L i ≤ dim (ker L i )
i=1
( ∏l
)
and so if you can find a linearly independent set of vectors in ker
i=1 L i of size
i=1
l∑
dim (ker L i ) ,
i=1
then it must be a basis for ker
( ∏l
i=1 L i
)
. This is discussed below.
Definition 10.1.4 Let {V i } r i=1
be subspaces of V. Then
r∑
denotes all sums of the form ∑ r
i=1 v i where v i ∈ V i . If whenever
i=1
V i
r∑
v i =0,v i ∈ V i , (10.1)
i=1
it follows that v i =0for each i, then a special notation is used to denote ∑ r
i=1 V i. This
notation is
V 1 ⊕···⊕V r ,
and it is called a direct sum of subspaces.
Lemma 10.1.5 If V = V 1 ⊕···⊕V r and if β i = { }
v1, i ··· ,vm i i is a basis for Vi , then a
basis for V is {β 1 , ··· ,β r }.
Proof: Suppose ∑ r ∑ mi
i=1 j=1 c ijvj i =0. then since it is a direct sum, it follows for each i,
∑m i
j=1
c ij v i j =0
and now since { v i 1, ··· ,v i m i
}
is a basis, each cij =0.
Here is a useful lemma.
Lemma 10.1.6 Let L i be in L (V,V ) and suppose for i ≠ j, L i L j = L j L i and also L i is
one to one on ker (L j ) whenever i ≠ j. Then
( p∏
)
ker L i =ker(L 1 ) ⊕ + ···+ ⊕ ker (L p )
i=1
Here ∏ p
i=1 L i is the product of all the linear transformations. A symbol like ∏ j≠i L j is the
product of all of them but L i .