Unformatted text preview: a simple eigenvalue is always semisimple, but not conversely. IG
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,
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
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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)
insures the existence of a unitary U such that U∗ An×n U =
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
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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