1
Proposition: Let X be a vector space and L ∈ L(X, Y ). Prove that
dim(X) = dim(nullL) + dim(L(X))
Proof: nullL is a subspace of X. As such it has a base, say V . If |V | = ∞, there is nothing to
prove. Similarly, if spanV = X there is nothing to prove.1 So let |V | < ∞ and dim(nullL) <
dim(X). We can extend V to V̄ ⊃ V a base for X. Define the set W := {L(v) : v ∈ V̄ \ V }.
Claim: W is base for L(X).
Subproof. First we want to show spanW = L(X). The inclusion ⊆ is trivial. For the other
inclusion, let y ∈ L(X). Then there exists x ∈ X s.t. L(x) = y. Being P
V̄ a base for X,
there exists a linear combination of element of V that equals x, i.e. x = n λi vi for some
{λi , vi }ni=1 ⊂ R × V̄ . Then we have




!
!
n
X
X
X
X
X
y = L(x) = L
λi vi = L 
λi vi +
λi vi  = L
λi vi + L 
λi vi 
vi ∈V
Notice that
P
vi ∈V
λi vi ∈ nullL hence L(

vi ∈V
vi ∈V̄ \V
P
vi ∈V
vi ∈V̄ \V
λi vi ) = 0. So we are left with

X
y = L
λi vi  =
vi ∈V̄ \V
X
vi ∈V̄ \V
λi L(vi ) =
X
λi wi .
w∈W
Thus, we have found a linear combination of elements of W that generates y. Since y was
arbitrary, spanW = L(X).
Second, we want to show that W is linearly independent. Suppose
not, i.e. suppose you can
P
find a combination with non zero coefficients such that 0 = w∈W λi wi . It follows that,


X
X
X
λi L(vi ) = L 
λi vi 
λi wi =
0=
w∈W
vi ∈V̄ \V
vi ∈V̄ \V
P
or that vi ∈V̄ \V λi vi ∈ nullL. Since V spans nullL, there exists a combination of elements of V
P
P
P
that generates vi ∈V̄ \V λi vi . That is for some {λ0i , vi0 }ni=1 ⊂ R×V , v0 ∈V λ0i vi0 = vi ∈V̄ \V λi vi .
i
To reach a contradiction, simply notice that we have found a {(λ0i , λ), (vi0 , vi )}ni=1 ⊂ R × V̄
such that
X
X
λi vi = 0.
λ0i vi0 −
vi0 ∈V
vi ∈V̄ \V
Since at least some of these λi are non zero (by our contrapositive assumption), we can argue
that V̄ is not independent. But V̄ was a base for X, hence it must be linearly independent,
a contradiction.
1
In the first case, this is because V ⊂ X implies ∞ = dim(nullL) ≤ dim(X). In the latter, if spanV = X,
then X = nullL and L(X) = {0}.