5.2. THE MATRIX OF A LINEAR TRANSFORMATION 103

where δ ik = 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, also rank one transformations.

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

Lvr =m

∑j=1

d jrw j

Now consider the following sum of dyadics. ∑mj=1 ∑

ni=1 d jiw jvi. Apply this to vr. This yields

m

∑j=1

n

∑i=1

d jiw jvi (vr) =m

∑j=1

n

∑i=1

d jiw jδ ir =m

∑j=1

d jrw j = Lvr (5.1)

Therefore, L = ∑mj=1 ∑

ni=1 d jiw jvi showing the span of the dyadics is all of L (V,W ) .

Now consider whether these special linear transformations are a linearly independentset. 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, thisshows dil = 0 for all i and l. Thus these linear transformations form a basis and this showsthat the dimension of L (V,W ) is mn as claimed because there are m choices for the wi andn choices for the v j. ■

Note that from 5.1, these coefficients which obtain L as a linear combination of thediadics are given by the equation

m

∑j=1

d jrw j = Lvr (5.2)

Thus Lvr is in the span of the w j.

5.2 The Matrix of a Linear TransformationIn order to do computations based on a linear transformation, we usually work with itsmatrix. This is what is described here.

Theorem 5.1.4 says that the rank one transformations defined there in terms of twobases, one for V and the other for W are a basis for L (V,W ) . Thus if A ∈L (V,W ) , thereare scalars Ai j such that

A =n

∑i=1

m

∑j=1

Ai jwiv j

Here we have 1≤ i≤ n and 1≤ j≤m. We can arrange these scalars in a rectangular shapeas follows. 

A11 A12 · · · A1(n−1) A1n

A21 A22 · · · A2(n−1) A2n...

......

...Am1 Am2 · · · Am(n−1) Amn

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