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⎠
⎝
0
0 It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] ⎛ ⎞
−4
⎜ 5⎟
⎟
⎜
⎜ 2⎟
v2 = ⎜
⎟,
⎜ 0⎟
⎠
⎝
3
0 ⎞
1
⎜ −2 ⎟
⎟
⎜
⎜ −2 ⎟
v3 = ⎜
⎟,
⎜ 0⎟
⎠
⎝
0
3 ⎛ ⎛ and D
E and {Bv1 , Bv2 , Bv3 } = {v1 , v2 , v3 }. Reducing [ b1 | b2 | v1 | v2 | v3 ] to
echelon form reveals that its basic columns are in positions one, two, and
three, so v1 is the needed extension vector. Therefore, the complete nested
basis for N (L) is
⎞
6
⎜ −6 ⎟
⎟
⎜
⎜ 0⎟
b1 = ⎜
⎟ ∈ S2 ,
⎜ 0⎟
⎠
⎝
−6
−6
⎛ ⎞
−5
⎜ 7⎟
⎟
⎜
⎜ 2⎟
b2 = ⎜
⎟ ∈ S1 ,
⎜ −2 ⎟
⎠
⎝
3
1 T
H ⎛ ⎞
2
⎜ −4 ⎟
⎟
⎜
⎜ −1 ⎟
b3 = ⎜
⎟ ∈ S0 .
⎜ 3⎟
⎠
⎝
0
0
⎛ and IG
R 4. Complete the process by building a Jordan chain on top of each bj ∈ Si
by solving Li xj = bj and by setting Jj = [Li xj | · · · | Lxj | xj ]. Since
x1 = e1 solves L2 x1 = b1 , we have J1 = [ L2 e1 | Le1 | e1 ]. Solving
Lx2 = b2 yields x2 = (−1, 0, 2, 0, 0, 0)T , so J2 = [ Lx2 | x2 ]. Finally,
J3 = [ b3 ]. Putting these chains together produces Y
P ⎛ 6
⎜ −6
⎜
⎜0
P = [ J1 | J2 | J3 ] = ⎜
⎜0
⎝
−6
−6 O
C 1
3
−2
2
−5
−3 ⎞
1 −5 −1
2
0
7
0 −4 ⎟
⎟
0
2
2 −1 ⎟
⎟.
0 −2
0
3⎟
⎠
0
3
0
0
0
1
0
0 It can be veriﬁed by direct multiplication that P−1 LP = N. It’s worthwhile to pay attention to how the results in this section translate into
the language of direct sum decompositions of invariant subspaces as discussed
in §4.9 (p. 259) and §5.9 (p. 383). For a linear nilpotent operator L of index
k deﬁned on a ﬁnite-dimensional vector space V , statement (7.7.6) on p. 579
means that V can be decomposed as a direct sum V = V1 ⊕ V2 ⊕ · · · ⊕ Vt , where
Vj = span(Jbj ) is the space spanned by a Jordan chain emanating from the
basis vector bj ∈ N (L) and where t = dim N (L). Furthermore, each Vj is an Copyright c 2000 SIAM Buy online from SIAM
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