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Unformatted text preview: , we know that λ = 0 is the only eigenvalue, so the set of eigenvectors of L is N (L) (excluding the zero vector of course). Realizing that L is not diagonalizable is equivalent to realizing that L does not possess a complete linearly independent set of eigenvectors or, equivalently, dim N (L) < n. As in the 3 × 3 example above, the strategy for building a similarity transformation P that reduces L to a simple triangular form is as follows. (1) Construct a somewhat special basis B for N (L). IG R Y P O C (2) Extend B to a basis for C n by building Jordan chains on top of the eigenvectors in B . To accomplish (1), consider the subspaces deﬁned by Mi = R Li ∩ N (L) for i = 0, 1, . . . , k, (7.7.1) and notice (Exercise 7.7.4) that these subspaces are nested as 0 = Mk ⊆ Mk−1 ⊆ Mk−2 ⊆ · · · ⊆ M1 ⊆ M0 = N (L). Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 576 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] Use these nested spaces to construct a basis for N (L) = M0 by starting with any basis Sk−1 for Mk−1 and by sequentially extending Sk−1 with additional sets Sk−2 , Sk−3 , . . . , S0 such that Sk−1 ∪ Sk−2 is a basis for Mk−2 , Sk−1 ∪ Sk−2 ∪ Sk−3 is a basis for Mk−3 , etc. In general, Si is a set of vectors that extends Sk−1 ∪ Sk−2 ∪ · · · ∪ Si−1 to a basis for Mi . Figure 7.7.1 is a heuristic diagram depicting an example of k = 5 nested subspaces Mi along with some typical extension sets Si that combine to form a basis for N (L). D E T H Figure 7.7.1 Now extend the basis B = Sk−1 ∪ Sk−2 ∪ · · · ∪ S0 = {b1 , b2 , . . . , bt } for N (L) to a basis for C n by building Jordan chains on top of each b ∈ B. If b ∈ Si , then there exists a vector x such that Li x = b because each b ∈ Si belongs to Mi = R Li...
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