148 CHAPTER 6. DIRECT SUMS AND BLOCK DIAGONAL MATRICES

(b) For each x ̸= 0, there is such a β x and let V1 ≡ span(β x1

). Explain why N :

V1→V1.

(c) Let N1 be the restriction of N to V1. Find the matrix of N1 with respect to theordered basis

{Nm−1x1, · · · ,Nx1,x1

}. Note that we reverse the order of these

vectors. This is just the traditional way of doing it. Show this matrix is of theform

B≡

0 1 0

0. . .. . . 1

0 0

 (6.3)

15. ↑In the context of the above problems where

N p = 0,N ∈L (V,V )

and β x is defined as above, show that for each k ≤ p, if W is a subspace of ker(Nk)

which is invariant with respect to N meaning N (W )⊆W, then there are finitely manyyi ∈W such that

W = span(β y1

,β y2, · · · ,β ys

), some s

and{

β y1,β y2

, · · · ,β ys

}is linearly independent. This is called a cyclic basis. Hint:

If W ⊆ ker(N) , this is obviously true because in this case, β x = x for x ∈ ker(N).Now suppose the assertion is true for k < p and consider invariant W ⊆ ker

(Nk+1

).

Argue as follows:

(a) Explain why N (W ) is an invariant subspace of ker(Nk). Thus, by induction,

N (W ) = span(β x1

,β x2, · · · ,β xs

)where that in (·) is a basis.

(b) Let z ∈W so Nz = ∑si=1 ∑

ri−1j=0 ai jN jx j. Let y j ∈W such that Ny j = x j. Explain

why

N

(z−

s

∑i=1

ri−1

∑j=0

ai jN jyi

)= 0

where the length of β xiis ri. Explain why there is an eigenvector y0 such that

z =s

∑i=1

ri−1

∑j=0

ai jN jyi + y0

(c) Note that β y0= y0. Explain why

span(

β y0,β y1

, · · · ,β ys

)⊇W

Then explain why{

β y0,β y1

, · · · ,β ys

}is linearly independent. Hint: If

s

∑i=1

ri−1

∑j=0

ai jN jyi +by0 = 0,

Do N to both sides and use induction to conclude all ai j = 0.