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# T h proof of 724 the result of exercise 5914

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Unformatted text preview: a simple eigenvalue is always semisimple, but not conversely. IG R Y P O C The algebraic and geometric multiplicity need not agree. For example, the nilpotent matrix A = 0 1 in (7.2.1) has only one distinct eigenvalue, λ = 0, 00 that is repeated twice, so alg multA (0) = 2. But dim N (A − 0I) = dim N (A) = 1 =⇒ geo multA (0) = 1. In other words, there is only one linearly independent eigenvector associated with λ = 0 even though λ = 0 is repeated twice as an eigenvalue. Example 7.2.3 shows that geo multA (λ) < alg multA (λ) is possible. However, the inequality can never go in the reverse direction. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.2 Diagonalization by Similarity Transformations http://www.amazon.com/exec/obidos/ASIN/0898714540 511 Multiplicity Inequality For every A ∈ C n×n , and for each λ ∈ σ (A), It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] geo multA (λ) ≤ alg multA (λ) . (7.2.2) D E Proof. Suppose alg multA (λ) = k. Schur’s triangularization theorem (p. 508) T11 T12 , insures the existence of a unitary U such that U∗ An×n U = 0 T22 where T11 is a k × k upper-triangular matrix whose diagonal entries are equal / to λ, and T22 is an n − k × n − k upper-triangular matrix with λ ∈ σ (T22 ) . Consequently, T22 − λI is nonsingular, so rank (A − λI) = rank (U∗ (A − λI)U) = rank T H T11 − λI 0 IG R ≥ rank (T22 − λI) = n − k. T12 T22 − λI The inequality follows from the fact that the rank of a matrix is at least as great as the rank of any submatrix—recall the result on p. 215. Therefore, Y P alg multA (λ) = k ≥ n − rank (A − λI) = dim N (A − λI) = geo multA (λ) . Determining whether or not An×n is diagonalizable is equivalent to determining whether or not A has a complete linearly independent set of eigenvectors, and this can be done if you are willing and able to compute all of the eigenvalues and eigenvectors for A. But this brute force approach can be a monument...
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