where V,W,U are all finite dimensional
vector spaces. Then it is interesting to consider ker
(BA )
. The following theorem of Sylvester is a
very useful and important result.
Theorem 9.1.3Let A ∈ℒ
(V, W )
and B ∈ℒ
(W, U)
where V,W,U are all vector spaces over afield F. Suppose also that ker
(A)
and A
(ker(BA ))
are finite dimensional subspaces.Then
dim (ker(BA )) ≤ dim (ker(B))+ dim (ker(A)).
Equality holds if and only if A
(ker(BA ))
= ker
(B )
.
Proof:If x ∈ ker
(BA )
, then Ax ∈ ker
(B)
and so A
(ker(BA ))
⊆ ker
(B)
. The following
picture may help.
PICT
Now let
{x1,⋅⋅⋅,xn}
be a basis of ker
(A)
and let
{Ay1, ⋅⋅⋅,Aym }
be a basis for A
(ker(BA ))
.
Take any z ∈ ker
(BA )
. Then Az = ∑_{i=1}^{m}a_{i}Ay_{i} and so
( m )
A z − ∑ a y = 0
i=1 i i
which means z −∑_{i=1}^{m}a_{i}y_{i}∈ ker
(A)
and so there are scalars b_{i} such that
∑m ∑n
z − aiyi = bixi.
i=1 j=1
It follows span
(x1,⋅⋅⋅,xn,y1,⋅⋅⋅,ym )
⊇ ker
(BA )
and so by the first part, (See the
picture.)
dim(ker(BA )) ≤ n+ m ≤ dim (ker (A )) + dim (ker(B ))
Now
{x1,⋅⋅⋅,xn,y1,⋅⋅⋅,ym}
is linearly independent because if
∑ ∑
aixi + bjyj = 0
i j
then you could do A to both sides and conclude that ∑_{j}b_{j}Ay_{j} = 0 which requires
that each b_{j} = 0. Then it follows that each a_{i} = 0 also because it implies ∑_{i}a_{i}x_{i} = 0.
Thus
{x1,⋅⋅⋅,xn,y1,⋅⋅⋅,ym}
is a basis for ker
(BA )
. Then A
(ker(BA ))
= ker
(B)
if and only if m = dim
(ker(B ))
if and only
if
dim(ker(BA )) = m + n = dim (ker (B ))+ dim (ker(A )).■
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
∏_{i=1}^{l}L_{i} all in ℒ
(V,V )
. Then
( )
∏l ∑l
dim ker Li ≤ dim (kerLi).
i=1 i=1
Definition 9.1.4Let
{V }
i
_{i=1}^{r}be subspaces of V. Then
∑r
Vi ≡ V1 + ⋅⋅⋅+ Vr
i=1
denotes all sums of the form∑_{i=1}^{r}v_{i}where v_{i}∈ V_{i}. If whenever
∑r
vi = 0,vi ∈ Vi, (9.1)
i=1
(9.1)
it follows that v_{i} = 0 for each i, then a special notation is used to denote∑_{i=1}^{r}V_{i}. This notationis
V1 ⊕ ⋅⋅⋅⊕ Vr,
and it is called a direct sumof subspaces.
Now here is a useful lemma which is likely already understood.
Lemma 9.1.5Let L ∈ ℒ
(V,W )
where V,W are n dimensional vector spaces. Then Lis one to one, if and only if L is also onto. In fact, if
{v1,⋅⋅⋅,vn}
is a basis, then so is
{Lv1,⋅⋅⋅,Lvn}
.
Proof:Let
{v1,⋅⋅⋅,vn}
be a basis for V . Then I claim that
{Lv1,⋅⋅⋅,Lvn}
is a basis for W.
First of all, I show
{Lv1,⋅⋅⋅,Lvn}
is linearly independent. Suppose
∑n
ckLvk = 0.
k=1
Then
( n )
∑
L ckvk = 0
k=1
and since L is one to one, it follows
n∑
ckvk = 0
k=1
which implies each c_{k} = 0. Therefore,
{Lv1,⋅⋅⋅,Lvn }
is linearly independent. If there exists w not
in the span of these vectors, then by Lemma 7.2.10,
{Lv1,⋅⋅⋅,Lvn,w }
would be independent and
this contradicts the exchange theorem, Theorem 7.2.4 because it would be a linearly independent
set having more vectors than the spanning set
{v1,⋅⋅⋅,vn}
.
Conversely, suppose L is onto. Then there exists a basis for W which is of the form
{Lv1,⋅⋅⋅,Lvn}
. It follows that
{v1,⋅⋅⋅,vn}
is linearly independent. Hence it is a basis for V by
similar reasoning to the above. Then if Lx = 0, it follows that there are scalars c_{i} such that
x = ∑_{i}c_{i}v_{i} and consequently 0 = Lx = ∑_{i}c_{i}Lv_{i}. Therefore, each c_{i} = 0 and so x = 0 also. Thus
L is one to one. ■
Lemma 9.1.6If V = V_{1}⊕
⋅⋅⋅
⊕V_{r}and if β_{i} =
{ }
vi1,⋅⋅⋅,vimi
is abasis for V_{i}, then a basis forV is
{β1,⋅⋅⋅,βr}
. Thus
r
dim (V ) = ∑ dim(Vi).
i=1
Proof: Suppose ∑_{i=1}^{r}∑_{j=1}^{mi}c_{ij}v_{j}^{i} = 0. then since it is a direct sum, it follows for each
i,
∑mi i
cijvj = 0
j=1
and now since
{ i i }
v1,⋅⋅⋅,vmi
is a basis, each c_{ij} = 0. ■
Here is a fundamental lemma.
Lemma 9.1.7Let L_{i}be in ℒ
(V,V)
and suppose for i≠j,L_{i}L_{j} = L_{j}L_{i}andalso L_{i}is one to oneon ker
(L )
j
whenever i≠j. Then
( )
p∏
ker Li = ker(L1)⊕ + ⋅⋅⋅+ ⊕ker(Lp)
i=1
Here∏_{i=1}^{p}L_{i}is the product of all the linear transformations.
Proof :Note that since the operators commute, L_{j} : ker
(L )
i
→ ker
(L )
i
. Here is why. If
L_{i}y = 0 so that y ∈ ker
(L )
i
, then
LiLjy = LjLiy = Lj0 = 0
and so L_{j} : ker
(Li)
↦→
ker
(Li)
. Next observe that it is obvious that, since the operators
commute,
( )
∑p p∏
ker(Lp) ⊆ ker Li
i=1 i=1
Next, why is ∑_{i} ker
(Lp)
= ker
(L1)
⊕
⋅⋅⋅
⊕ ker
(Lp)
? Suppose
∑p
vi = 0,vi ∈ ker(Li),
i=1
but some v_{i}≠0. Then do ∏_{j≠i}L_{j} to both sides. Since the linear transformations commute, this
results in
∏
Lj(vi) = 0
j⁄=i
which contradicts the assumption that these L_{j} are one to one on ker
(Li)
and the observation
that they map ker
(Li)
to ker
(Li)
. Thus if
∑
vi = 0,vi ∈ ker(Li)
i
then each v_{i} = 0. It follows that
( )
p∏
ker(L1)⊕ + ⋅⋅⋅+ ⊕ ker(Lp) ⊆ ker Li (*)
i=1
(*)
From Sylvester’s theorem and the observation about direct sums in Lemma 9.1.6,
∑p
dim(ker(Li)) = dim (ker (L1) ⊕ +⋅⋅⋅+ ⊕ ker (Lp))
i=1 ( ( p )) p
∏ ∑
≤ dim ker i=1 Li ≤ i=1dim (ker(Li))
which implies all these are equal. 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 so the dimension of V would then be strictly larger than the dimension of W. It
follows from * that