Linear Algebra, Theory And Applications, 2012a

226 LINEAR TRANSFORMATIONS
Lemma 9.2.2 Let V and W be vector spaces and suppose {v 1 , ··· ,v n } is a basis for V.
Then if L : V → W is given by Lv k = w k ∈ W and
( n
)
∑
n∑
n∑
L a k v k ≡ a k Lv k = a k w k
k=1
k=1
then L is well defined and is in L (V,W) . Also, if L, M are two linear transformations such
that Lv 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 ∑ n
k=1 a kv k is an arbitrary
vector of V, ( n
) (
∑
n∑
n∑
n
)
∑
L a k v k = a k Lv k = a k Mv k = M a k v k
k=1
k=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
doestoabasis.
Theorem 9.2.3 Let V and W be finite dimensional linear spaces of dimension n and m
respectively Then dim (L (V,W)) = mn.
Proof: Let two sets of bases be
k=1
k=1
{v 1 , ··· ,v n } and {w 1 , ··· ,w m }
for V and W respectively. Using Lemma 9.2.2, let w i v j ∈L(V,W) be the linear transformation
defined on the basis, {v 1 , ··· ,v n }, by
w i v k (v j ) ≡ w i δ jk
where δ ik =1ifi = 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
m∑
Lv r = d jr w j
Now consider the following sum of dyadics.
Apply this to v r . This yields
j=1 i=1
m∑
j=1 i=1
j=1
n∑
d ji w j v i
j=1 i=1
k=1
m∑ n∑
m∑ n∑
m∑
d ji w j v i (v r )= d ji w j δ ir = d jr w i = Lv r
Therefore, L = ∑ m ∑ n
j=1 i=1 d jiw j v i showing the span of the dyadics is all of L (V,W) .
Now consider whether these dyadics form a linearly independent set. Suppose
∑
d ik w i v k = 0.
i,k
j=1