0 1 the above example suggests that determining the

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Unformatted text preview: ⎠ ⎝ 0 0 It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu ⎛ ⎞ −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 http://www.ec-securehost.com/SIAM/ot71.html Buy...
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This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

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