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 ! $ = ' $ I l eigenvectors where ' is called an eigenvalue . eigenvalues eigeipairs If $ is an eigenvector, then α$ is also an eigenvector. Therefore, we will usually seek for normalized eigenvectors , so that -- ay I - Hau ) III $ = 1 11 × 112--1 Aku ) - Note: When using Python, numpy.linalg.eig will normalize using p=2 norm.

How do we find eigenvalues? Linear algebra approach: = I ! $ = ' $ ! − ' , $ = & 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 of a polynomial of degree 5 or higher) ⟹ Approximate the eigenvalues numerically !

. D . → rank CA > = I columns of A- are L J Example A- is singular matrix - KI ) delta - O - ! = 2 1 &'( 2 − * 1 2 − * = 0 4 2 4 - - 212 ) X t k p CX ) = ( 2 - XJ → 4 - 4--0 - 4=0 height :* 12 - 42--0 eigenvectors " x. ⇒ ⇒ . ÷ ,f¥i¥¥÷¥¥ - XI ) x CA = 0 to ) to :o) . * : : - (d) lx I - 2L . I . eigenvectors

Diagonalizable Matrices A "×" matrix ! with " linearly independent eigenvectors 3 is said to be A X , H4 → A4=h# diagonalizable . a Hii FEEL ! 3 ! = ' " 3 ! , iii. ! 3 # = ' $ 3 # , … ! 3 % = ' & 3 % , na - * ÷ ÷¥¥¥÷s¥÷÷÷¥a " " K . .

II ::÷÷ &'( 2 − * 1 ! = 2 1 Example 2 − * = 0 4 4 2 Solution of characteristic polynomial gives: ' " = 4, ' $ = 0 To get the eigenvectors, we solve: ! $ = ' $ o Ipx :b - ! 2 − (4) 1 = 0 0 = 1 line . - " 4 2 − (4) 0 2 . :c :) → * to O I - ! 2 − (0) 1 0 = −1 = 0 - " 2 4 2 − (0) 0 t.tt#E/e.t:::I.o:::Igtisaias . o ) I. =L " - O O

- I -361=63 # O det (A) = 27 Example → NOT SINGULAR The eigenvalues of the matrix: ( A - XI ) x - o - l : " : I :) ! = 3 −18 2 −9 too ) are ' " = ' $ = −3 . K one eigenvector Select the incorrect statement: 261 - Geez → Jet - 1862=0 → False 6 U , A) Matrix ! is diagonalizable → True B) The matrix ! has only one eigenvalue with multiplicity 2 C) Matrix ! has only one linearly independent eigenvector - → True D) Matrix ! is not singular - → True onlyoneeigenecfor A=¥

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.

- ai A !×! symmetric matrix # with ! distinct eigenvalues is diagonalizable. Mio → Av= µ o Suppose $ , % and &, ( are eigenpairs of # → vector - $ % = #% -- Ay # It & ( = #( LI → scalars - XE.nu Ak - ftp.u-XEEorghogond vectors A- symmetric → ftp.ye-xv.y ^ II - AFI AIA my .ie ⇒ le/ = = - A) LEE ) - o ( M - ¥0

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? A , nxn → Ui , Ue - , Un → L - I . - - , Assume that # is diagonalizable (i.e., it has * linearly independent eigenvectors % ). We can propose a vector + which is a linear combination of these /AUi=XiUT eigenvectors: =/ + = , ! % ! + , " % " + ⋯ + , # % # . t Adn Un #If= Aau , t A talk t . . - tank kn = Idek , tazz Kz t - - - Assume : - 71 Xml 1412112171 Xsl - -

Xo = - I Power Iteration ¥ M Our goal is to find an eigenvector % $ of #. We will use an iterative process, converge - where we start with an initial vector, where here we assume that it can be → diagonal ie ab le D written as a linear combination of the eigenvectors of # . all L . I . Ui + % = , ! % ! + , " % " + ⋯ + , # % # = I , 4947 faction n Un t an X A- Io = 9 X , U , t da da Uz t - . . . t Adn X n Un = A a , X , U , t A da Xa Uz t A- Ii . . ) - t an X n ( x n un ) take ( we a) t - tan in un = If Ethier - = a , X , H , Ui - = a , Xiu , t k Xi uz t . - . - t in XP Un = Is , t da XP Uz t It ¥2 = a , XP U - - - t an X nk Un A. x ÷ , = I r = di X Y U i t da Tf Uz t - -

- t an Un = All , the Uz t Xo Power Iteration - - Anxn A OX 2 2 ' $ ' & $ 2 = ' " 2 < " 3 " + < $ 3 $ + ⋯ + < & 3 & ' " ' " - dometnaut Assume that < " ≠ 0 , the term < " 3 " dominates the others when ? is - I Xnl 14131121 > last very large. - - 2 Since |' " > |' $ , we have 3 ! ≪ 1 when ? is large 3 " Hence, as ? increases, $ 2 converges to a multiple of the first eigenvector 3 " , i.e., Tp " 't ' In - →④ %y , k → a → le Hxkll - grow ↳ large !