The linear transformations can be considered as a vector space as described next.
Definition 5.1.1 Given L,M ∈ℒ

For α a scalar and L ∈ℒ

You should verify that all the axioms of a vector space hold for ℒ
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 5.1.2 Let V and W be vector spaces and suppose {v_{1},

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

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.
Example 5.1.3 A basis for ℝ^{2} is

Suppose T is a linear transformation which satisfies

Find T
Theorem 5.1.4 Let V and W be finite dimensional linear spaces of dimension n and m respectively. Then dim
Proof: Let two sets of bases be

for V and W respectively. Using Lemma 5.1.2, let w_{i}v_{j} ∈ℒ

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

Now consider the following sum of dyadics.

Apply this to v_{r}. This yields
 (5.1) 
Therefore, L = ∑ _{j=1}^{m} ∑ _{i=1}^{n}d_{ji}w_{j}v_{i} showing the span of the dyadics is all of ℒ
Now consider whether these dyadics form a linearly independent set. Suppose

Are all the scalars d_{ik} equal to 0?

and so, since {w_{1},
Note that from 5.1, these coefficients which obtain L as a linear combination of the diadics are given by the equation
 (5.2) 
Thus Lv_{r} is in the span of the w_{j} and the scalars in the linear combination are d_{1r},d_{2r},
class=”left” align=”middle”(V,W) As A Vector Space5.2. THE MATRIX OF A LINEAR TRANSFORMATION