18.3. EIGENVALUES AND EIGENVECTORS OF LINEAR TRANSFORMATIONS427

then both L and M are equal on the basis for V . Therefore, L = M. Given such an m× nmatrix C = (ci j) , use the above formula to define L. Thus θ is a one to one and onto mapfrom L (V,W ) to the space of m× n matrices Mmn. It is also clear that θ is a linear mapbecause if θL =C and θM = D and a,b scalars,

(aL+bM)vk =m

∑i=1

acikvi +m

∑i=1

bdikvi =m

∑i=1

(acik +bdik)vi

and so θ (aL+bM) = aC+bD.Obviously this space of matrices is of dimension mn, a basis consisting of Ei j the matrix

which has a 1 in the i jth position and zeros elsewhere. Therefore,{

θ−1Ei j

}i. j is a basis for

L (V,W ) by the above lemma. ■

18.3 Eigenvalues And Eigenvectors Of Linear Transfor-mations

Here is a very useful theorem due to Sylvester.

Theorem 18.3.1 Let A ∈L (V,W ) and B ∈L (W,U) where V,W,U are all vector spacesover a field F. Suppose also that ker(A) and A(ker(BA)) are finite dimensional subspaces.Then

dim(ker(BA))≤ dim(ker(B))+dim(ker(A)) .

Proof: If x ∈ ker(BA) , then Ax ∈ ker(B) and so A(ker(BA))⊆ ker(B) . The followingpicture may help.

ker(B)

A(ker(BA))

ker(BA)

ker(A)A

Now let {x1, · · · ,xn} be a basis of ker(A) and let {Ay1, · · · ,Aym} be a basis for

A(ker(BA)) .

Take any z ∈ ker(BA) . Then Az = ∑mi=1 aiAyi and so

A

(z−

m

∑i=1

aiyi

)= 0

which means z−∑mi=1 aiyi ∈ ker(A) and so there are scalars bi such that

z−m

∑i=1

aiyi =n

∑j=1

bixi.

It follows span(x1, · · · ,xn,y1, · · · ,ym)⊇ ker(BA) and so by the first part, (See the picture.)

dim(ker(BA))≤ n+m≤ dim(ker(A))+dim(ker(B)) ■