216 CHAPTER 8. LINEAR TRANSFORMATIONS

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 8.2.3 Let V and W be finite dimensional linear spaces 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 wivj ∈ L (V,W ) be the linear transfor-mation defined on the basis, {v1, · · · , vn}, by

wivk (vj) ≡ wiδjk

where δik = 1 if i = k and 0 if i ̸= k. I will show that L ∈ L (V,W ) is a linear combinationof these special linear transformations called dyadics.

Then let L ∈ L (V,W ). Since {w1, · · · , wm} is a basis, there exist constants, djk suchthat

Lvr =

m∑j=1

djrwj

Now consider the following sum of dyadics.

m∑j=1

n∑i=1

djiwjvi

Apply this to vr. This yields

m∑j=1

n∑i=1

djiwjvi (vr) =

m∑j=1

n∑i=1

djiwjδir =

m∑j=1

djrwi = Lvr

Therefore, L =∑m

j=1

∑ni=1 djiwjvi showing the span of the dyadics is all of L (V,W ) .

Now consider whether these dyadics form a linearly independent set. Suppose∑i,k

dikwivk = 0.

Are all the scalars dik equal to 0?

0 =∑i,k

dikwivk (vl) =

m∑i=1

dilwi

and so, since {w1, · · · , wm} is a basis, dil = 0 for each i = 1, · · · ,m. Since l is arbitrary,this shows dil = 0 for all i and l. Thus these linear transformations form a basis and thisshows that the dimension of L (V,W ) is mn as claimed because there are m choices for thewi and n choices for the vj . ■