23-diagonalization

# 23-diagonalization - DIAGONALIZATION Math 21b, O.Knill...

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Unformatted text preview: DIAGONALIZATION Math 21b, O.Knill HOMEWORK: Section 7.4: 14,20,26,36,54,58*,56* SUMMARY. A n × n matrix, A~v = λ~v with eigenvalue λ and eigenvector ~v . The eigenvalues are the roots of the characteristic polynomial f A ( λ ) = det( λ- A ) = λ n- tr( A ) λ n- 1 + ... + (- 1) n det( A ). The eigenvectors to the eigenvalue λ are in ker( λ- A ). The number of times, an eigenvalue λ occurs in the full list of n roots of f A ( λ ) is called algebraic multiplicity. It is bigger or equal than the geometric multiplicity: dim(ker( λ- A ). EXAMPLE. The eigenvalues of a b c d are λ ± = T/ 2 + p T 2 / 4- D , where T = a + d is the trace and D = ad- bc is the determinant of A . If c 6 = 0, the eigenvectors are v ± = λ ±- d c . If c = 0, then a, d are eigenvalues to the eigenvectors a and- b a- d . If a = d , then the second eigenvector is parallel to the first and the geometric multiplicity of the eigenvalue a = d is 1. EIGENBASIS. If A has n different eigenvalues, then A has an eigenbasis, consisting of eigenvectors of A . DIAGONALIZATION. How does the matrix A look in an eigenbasis? If S is the matrix with the eigenvectors as columns, then we know B = S- 1 AS . We have S~ e i = ~v i and AS~ e i = λ i ~v i we know S- 1 AS~ e i = λ i ~ e i . Therefore, B is diagonal with diagonal entries λ i . EXAMPLE. A = 2 3 1 2 has the eigenvalues λ 1 = 2 + √ 3 with eigenvector ~v 1 = [ √ 3 , 1] and the eigenvalues λ 2 = 2- √ 3 with eigenvector ~v 2 = [- √ 3 , 1]. Form S = √ 3- √ 3 1 1 and check S- 1 AS = D is diagonal....
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## This note was uploaded on 04/06/2008 for the course MATH 21B taught by Professor Judson during the Spring '03 term at Harvard.

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