**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
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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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