Linear Algebra, Theory And Applications, 2012a

10.1. A THEOREM OF SYLVESTER, DIRECT SUMS 247
Proof: Note that since the operators commute, L j :ker(L i ) → ker (L i ). Here is why. If
L i y =0sothaty ∈ ker (L i ) , then
and so L j :ker(L i ) → ker (L i ). Suppose
L i L j y = L j L i y = L j 0=0
p∑
v i =0,v i ∈ ker (L i ) ,
i=1
but some v i ≠0. Then do ∏ j≠i L j to both sides. Since the linear transformations commute,
this results in
∏
L j (v i )=0
j≠i
which contradicts the assumption that these L j are one to one and the observation that
they map ker (L i )toker(L i ). Thus if
∑
v i =0,v i ∈ ker (L i )
i
then each v i =0.
Suppose β i = { v i 1, ··· ,v i m i
}
is a basis for ker (Li ). Then from what was just shown and
Lemma 10.1.5, { β 1 , ··· ,β p
}
must be linearly independent and a basis for
ker (L 1 ) ⊕ + ···+ ⊕ ker (L p ) .
It is also clear that since these operators commute,
ker (L 1 ) ⊕ + ···+ ⊕ ker (L p ) ⊆ ker
( p∏
i=1
L i
)
Therefore, by Sylvester’s theorem and the above,
( ( p∏
))
p∑
dim ker L i ≤ dim (ker (L j ))
i=1
j=1
( ( p∏
))
=dim(ker(L 1 ) ⊕ + ···+ ⊕ ker (L p )) ≤ dim ker L i .
i=1
Now in general, if W is a subspace of V, a finite dimensional vector space and the two have
the same dimension, then W = V . This is because W has a basis and if v is not in the span
of this basis, then v adjoined to the basis of W would be a linearly independent set, yielding
a linearly independent set which has more vectors in it than a basis, a contradiction.
It follows that
( p∏
)
ker (L 1 ) ⊕ + ···+ ⊕ ker (L p )=ker L i
i=1