Eigenvalues and Eigenvectors Eigenvalue problem Let ! be an - - PowerPoint PPT Presentation

Eigenvalues and Eigenvectors

Eigenvalue problem

Let ! be an "×" matrix: $ ≠ & is an eigenvector of ! if there exists a scalar ' such that ! $ = ' $ where ' is called an eigenvalue. If $ is an eigenvector, then α$ is also an eigenvector. Therefore, we will usually seek for normalized eigenvectors, so that $ = 1 Note: When using Python, numpy.linalg.eig will normalize using p=2 norm.

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How do we find eigenvalues?

Linear algebra approach: ! $ = ' $ ! − ' , $ = & Therefore the matrix ! − ' , is singular ⟹ ./0 ! − ' , = 0 2 ' = ./0 ! − ' , is the characteristic polynomial of degree ". In most cases, there is no analytical formula for the eigenvalues of a matrix (Abel proved in 1824 that there can be no formula for the roots

• f a polynomial of degree 5 or higher) ⟹ Approximate the

eigenvalues numerically!

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Example

! = 2 1 4 2 &'( 2 − * 1 4 2 − * = 0

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Diagonalizable Matrices

A "×" matrix ! with " linearly independent eigenvectors 3 is said to be diagonalizable. ! 3! = '" 3!, ! 3# = '$ 3#, … ! 3% = '& 3%,

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Example

! = 2 1 4 2 &'( 2 − * 1 4 2 − * = 0

Solution of characteristic polynomial gives: '" = 4, '$ = 0 To get the eigenvectors, we solve: ! $ = ' $

2 − (4) 1 4 2 − (4)

• !
• "

= 0 0 = 1 2 2 − (0) 1 4 2 − (0)

• !
• "

= 0 0 = −1 2

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Example

The eigenvalues of the matrix: ! = 3 −18 2 −9 are '" = '$ = −3. Select the incorrect statement: A) Matrix ! is diagonalizable B) The matrix ! has only one eigenvalue with multiplicity 2 C) Matrix ! has only one linearly independent eigenvector D) Matrix ! is not singular det(A)= 27

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Let’s look back at diagonalization…

1) If a "×" matrix ! has " linearly independent eigenvectors $ then ! is diagonalizable, i.e., ! = :;:1! where the columns of : are the linearly independent normalized eigenvectors $ of ! (which guarantees that :1! exists) and ; is a diagonal matrix with the eigenvalues of !. 2) If a "×" matrix ! has less then " linearly independent eigenvectors, the matrix is called defective (and therefore not diagonalizable). 3) If a "×" symmetric matrix ! has " distinct eigenvalues then ! is diagonalizable.

A !×! symmetric matrix # with ! distinct eigenvalues is diagonalizable. Suppose $,% and &, ( are eigenpairs of # $ % = #% & ( = #(

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Some things to remember about eigenvalues:

• Eigenvalues can have zero value
• Eigenvalues can be negative
• Eigenvalues can be real or complex numbers
• A "×" real matrix can have complex eigenvalues
• The eigenvalues of a "×" matrix are not necessarily unique. In fact,

we can define the multiplicity of an eigenvalue.

• If a "×" matrix has " linearly independent eigenvectors, then the

matrix is diagonalizable

How can we get eigenvalues numerically?

Assume that # is diagonalizable (i.e., it has * linearly independent eigenvectors %). We can propose a vector + which is a linear combination of these eigenvectors: + = ,!%! + ,"%" + ⋯ + ,#%#

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• tankkn

Assume :

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Power Iteration

Our goal is to find an eigenvector %$ of #. We will use an iterative process, where we start with an initial vector, where here we assume that it can be written as a linear combination of the eigenvectors of #. +% = ,!%! + ,"%" + ⋯ + ,#%#

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→diagonalieable

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all L.I .

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Power Iteration

$2 = '" 2 <"3" + <$ '$ '"

2

3$ + ⋯ + <& '& '"

2

3& Assume that <" ≠ 0, the term <"3" dominates the others when ? is very large. Since |'" > |'$ , we have 3!

3" 2

≪ 1 when ? is large Hence, as ? increases, $2 converges to a multiple of the first eigenvector 3", i.e.,

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