You should verify that all the axioms of a vector space hold for ℒ

(V,W )

with the above
definitions of vector addition and scalar multiplication. What about the dimension of
ℒ

(V,W )

?

Before answering this question, here is a useful lemma. It gives a way to define linear
transformations and a way to tell when two of them are equal.

Lemma 8.2.2Let V and W be vector spaces and suppose {v_{1},

⋅⋅⋅

,v_{n}} is a basis for V. Then ifL : V → W is given by Lv_{k} = w_{k}∈ W and

( )
∑n ∑n ∑n
L akvk ≡ akLvk = akwk
k=1 k=1 k=1

then L is well defined and is in ℒ

(V,W )

. Also, if L,M are two linear transformations such thatLv_{k} = Mv_{k}for all k, then M = L.

Proof: L is well defined on V because, since {v_{1},

⋅⋅⋅

,v_{n}} is a basis, there is exactly one way to
write a given vector of V as a linear combination. Next, observe that L is obviously linear from
the definition. If L,M are equal on the basis, then if ∑_{k=1}^{n}a_{k}v_{k} is an arbitrary vector of
V,

( n ) n n ( n )
L ∑ a v = ∑ a Lv = ∑ a M v = M ∑ a v
k=1 k k k=1 k k k=1 k k k=1 k k

and so L = M because they give the same result for every vector in V . ■

The message is that when you define a linear transformation, it suffices to tell what it does to a
basis.

Theorem 8.2.3Let V and W be finite dimensional linearspaces of dimension n and mrespectively Then dim

(ℒ(V,W ))

= mn.

Proof:Let two sets of bases be

{v1,⋅⋅⋅,vn} and {w1,⋅⋅⋅,wm}

for V and W respectively. Using Lemma 8.2.2, let w_{i}v_{j}∈ℒ

(V,W )

be the linear transformation
defined on the basis, {v_{1},

⋅⋅⋅

,v_{n}}, by

wivk (vj) ≡ wiδjk

where δ_{ik} = 1 if i = k and 0 if i≠k. I will show that L ∈ℒ

(V,W )

is a linear combination of these
special linear transformations called dyadics.

Then let L ∈ℒ

(V,W )

. Since {w_{1},

⋅⋅⋅

,w_{m}} is a basis, there exist constants, d_{jk} such
that