6.6. EXERCISES 145
11. Now in this problem, the method for finding the special basis for a nilpotent trans-formation is given. Let V be a vector space and let N ∈L (V,V ) be nilpotent. Firstnote the only eigenvalue of N is 0. Why? (See Problem 8.) Let v1 be an eigenvector.Then {v1,v2, · · · ,vr} is called a chain based on v1 if Nvk+1 = vk for all k = 1,2, · · · ,rand v1 is an eigenvector so Nv1 = 0. It will be called a maximal chain if there is nosolution v, to the equation, Nv = vr. Now there will be a sequence of steps leadingto the desired basis.
(a) Show the vectors in any chain are linearly independent and for {v1,v2, · · · ,vr}a chain based on v1,
N : span(v1,v2, · · · ,vr) 7→ span(v1,v2, · · · ,vr) . (6.2)
Also if {v1,v2, · · · ,vr} is a chain, then r≤ n. Hint: If 0 = ∑ri=1 civi, and the last
nonzero scalar occurs at l, do Nl−1 to the sum and see what happens to cl .
(b) Consider the set of all chains based on eigenvectors. Since all have total lengthno larger than n it follows there exists one of maximal length,
{v1
1, · · · ,v1r1
}≡
B1. If span(B1) contains all eigenvectors of N, then stop. Otherwise, con-sider all chains based on eigenvectors not in span(B1) and pick one, B2 ≡{
v21, · · · ,v2
r2
}which is as long as possible. Thus r2 ≤ r1. If span(B1,B2) con-
tains all eigenvectors of N, stop. Otherwise, consider all chains based on eigen-vectors not in span(B1,B2) and pick one, B3 ≡
{v3
1, · · · ,v3r3
}such that r3 is as
large as possible. Continue this way. Thus rk ≥ rk+1. Then show that theabove process terminates with a finite list of chains {B1, · · · ,Bs} because forany k,{B1, · · · ,Bk} is linearly independent. Hint: From part a. you know thisis true if k = 1. Suppose true for k−1 and letting L(Bi) denote a linear combina-tion of vectors of Bi, suppose ∑
ki=1 L(Bi) = 0. Then we can assume L(Bk) ΜΈ= 0
by induction. Let vki be the last term in L(Bk) which has nonzero scalar. Now
act on the whole thing with Ni−1 to find vk1 as a linear combination of vectors
in {B1, · · · ,Bk−1} , a contradiction to the construction. You fill in the details.
(c) Suppose Nw = 0. (w is an eigenvector). Show that there exist scalars, ci suchthat w = ∑
si=1 civi
1. Recall that vi1 is the eigenvector in the ith chain on which
this chain is based. You know that w is a linear combination of the vectorsin {B1, · · · ,Bs} . This says that in fact it is a linear combination of the bottomvectors in the Bi. Hint: You know that w = ∑
si=1 L(Bi) . Let vs
i be the last inL(Bs) which has nonzero scalar. Suppose that i > 1. Now do Ni−1 to both sidesand obtain that vs
1 is in the span of {B1, · · · ,Bs−1} which is a contradiction.Hence i = 1 and so the only term of L(Bs) is one involving an eigenvector.Now do something similar to L(Bs−1) ,L(Bs−2) etc. You fill in details.
(d) If Nw = 0, then w ∈ span(B1, · · · ,Bs) . This was what was just shown. In fact,it was a particular linear combination involving the bases of the chains. What ifNkw= 0? Does it follow that w∈ span(B1, · · · ,Bs)? Show that if Nkw= 0, thenw ∈ span(B1, · · · ,Bs) . Hint: Say k is as small as possible such that Nkw = 0.Then you have Nk−1w is an eigenvector and so Nk−1w = ∑
si=1 civi
1 If Nk−1w isthe base of some chain Bi, then there is nothing to show. Otherwise, considerthe chain Nk−1w,Nk−2w, · · · ,w. It cannot be any longer than any of the chainsB1,B2, · · · ,Bs why? Therefore, vi
1 = Nk−1vik. Why is vi
k ∈ Bi? This is where you