426 CHAPTER 18. LINEAR TRANSFORMATIONS

Proof: L is well defined on V because, since {v1, · · · ,vn} is a basis, there is exactly oneway to write a given vector of V as a linear combination. Next, observe that L is obviouslylinear from the definition. If L,M are equal on the basis, then an arbitrary vector in V is ofthe form ∑

nk=1 akvk. Therefore,

L

(n

∑k=1

akvk

)=

n

∑k=1

akLvk =n

∑k=1

akMvk = M

(n

∑k=1

akvk

)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 itdoes to a basis.

Lemma 18.2.3 Suppose θ ∈L (X ,Y ) where X ,Y are vector spaces and θ is one to oneand onto. Then if {y1, · · · ,yn} is a basis for Y, it follows that{

θ−1y1, · · · ,θ−1yn

}is a basis for X.

Proof: Let xk = θ−1yk. If ∑k ckxk = 0, then

θ

(∑k

ckxk

)= ∑

kckθxk = ∑

kckyk = 0

and so, each ck = 0. Hence {x1, · · · ,xn} is independent. If x ∈ X , then θx ∈ Y so it equalsan expression of the form

n

∑k=1

ckyk = ∑k

ckθxk.

Hence

θ

(x−∑

kckxk

)= 0

and so, since θ is one to one, x−∑k ckxk = 0 which shows that {x1, · · · ,xn} also spans andis therefore, a basis. ■

Theorem 18.2.4 Let V and W be finite dimensional linear spaces of dimension n and mrespectively Then dim(L (V,W )) = mn.

Proof: Let the two sets of bases be

{v1, · · · ,vn} and {w1, · · · ,wm}

for X and Y respectively. Let L be a linear transformation. Then there are unique (since thew j form a basis) scalars ci j such that

Lvk =m

∑i=1

cikwi

Let C denote the m× n matrix whose i jth entry is ci j. Let θ be the mapping which takesL∈L (V,W ) to this matrix which is just defined. Then θ is one to one because if θL= θM,