80 CHAPTER 2. LINEAR TRANSFORMATIONS

and why there exist scalars ai such that

A (z − (a1y1 + · · ·+ amym)) = 0

and why it follows z − (a1y1 + · · ·+ amym) ∈ span {x1, · · · ,xn}. Now explain why

ker (BA) ⊆ span {x1, · · · ,xn,y1, · · · ,ym}

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

This important inequality is due to Sylvester. Show that equality holds if and only ifA(kerBA) = ker(B).

41. Generalize the result of the previous problem to any finite product of linear mappings.

42. If W ⊆ V for W,V two subspaces of Fn and if dim (W ) = dim (V ) , show W = V .

43. Let V be a subspace of Fnand let V1, · · · , Vm be subspaces, each contained in V . Then

V = V1 ⊕ · · · ⊕ Vm (2.37)

if every v ∈ V can be written in a unique way in the form

v = v1 + · · ·+ vm

where each vi ∈ Vi. This is called a direct sum. If this uniqueness condition does nothold, then one writes

V = V1 + · · ·+ Vm

and this symbol means all vectors of the form

v1 + · · ·+ vm, vj ∈ Vj for each j.

Show 2.37 is equivalent to saying that if

0 = v1 + · · ·+ vm, vj ∈ Vj for each j,

then each vj = 0. Next show that in the situation of 2.37, if βi ={ui1, · · · , uimi

}is a

basis for Vi, then {β1, · · · , βm} is a basis for V .

44. ↑Suppose you have finitely many linear mappings L1, L2, · · · , Lm which map V to Vwhere V is a subspace of Fn and suppose they commute. That is, LiLj = LjLi for alli, j. Also suppose Lk is one to one on ker (Lj) whenever j ̸= k. Letting P denote theproduct of these linear transformations, P = L1L2 · · ·Lm, first show

ker (L1) + · · ·+ ker (Lm) ⊆ ker (P )

Next show Lj : ker (Li) → ker (Li) . Then show

ker (L1) + · · ·+ ker (Lm) = ker (L1)⊕ · · · ⊕ ker (Lm) .

Using Sylvester’s theorem, and the result of Problem 42, show

ker (P ) = ker (L1)⊕ · · · ⊕ ker (Lm)

Hint: By Sylvester’s theorem and the above problem,

dim (ker (P )) ≤∑i

dim (ker (Li))

= dim (ker (L1)⊕ · · · ⊕ ker (Lm)) ≤ dim (ker (P ))

Now consider Problem 42.