18.3. EIGENVALUES AND EIGENVECTORS OF LINEAR TRANSFORMATIONS429

Proof: Note that since the operators commute, L j : ker(Li) 7→ ker(Li). Here is why. IfLiy = 0 so that y ∈ ker(Li) , then

LiL jy = L jLiy = L j0 = 0

and so L j : ker(Li) 7→ ker(Li). Next observe that it is obvious that, since the operatorscommute,

p

∑i=1

ker(Lp)⊆ ker

(p

∏i=1

Li

)Suppose

p

∑i=1

vi = 0, vi ∈ ker(Li) ,

but some vi ̸= 0. Then do ∏ j ̸=i L j to both sides. Since the linear transformations commute,this results in

∏j ̸=i

L jvi = 0

which contradicts the assumption that these L j are one to one and the observation that theymap ker(Li) to ker(Li). Thus if

∑i

vi = 0, vi ∈ ker(Li)

then each vi = 0. It follows that

ker(L1)⊕+ · · ·+⊕ker(Lp)⊆ ker

(p

∏i=1

Li

)(*)

From Sylvester’s theorem and the observation about direct sums in Lemma 18.3.3,

p

∑i=1

dim(ker(Li)) = dim(ker(L1)⊕+ · · ·+⊕ker(Lp))

≤ dim

(ker

(p

∏i=1

Li

))≤

p

∑i=1

dim(ker(Li))

which implies all these are equal. Now in general, if W is a subspace of V, a finite dimen-sional vector space and the two have the same dimension, then W =V . This is because Whas a basis and if v is not in the span of this basis, then v adjoined to the basis of W wouldbe a linearly independent set so the dimension of V would then be strictly larger than thedimension of W . It follows from * that

ker(L1)⊕+ · · ·+⊕ker(Lp) = ker

(p

∏i=1

Li

)■

Here is a situation in which the above holds. ker(A−λ iI)r is sometimes called a gen-

eralized eigenspace. The following is an important result on generalized eigenspaces.