Diagonalization Marco Chiarandini Department of Mathematics & - PowerPoint PPT Presentation

DM559 Linear and Integer Programming Lecture 10 Diagonalization Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

Diagonalization Outline Applications 1. Diagonalization 2. Applications 2

Diagonalization Outline Applications 1. Diagonalization 2. Applications 3

Diagonalization Eigenvalues and Eigenvectors Applications (All matrices in this lecture are square n × n matrices and all vectors in R n ) Definition Let A be a square matrix. • The number λ is said to be an eigenvalue of A if for some non-zero vector x , A x = λ x • Any non-zero vector x for which this equation holds is called eigenvector for eigenvalue λ or eigenvector of A corresponding to eigenvalue λ 4

Diagonalization Finding Eigenvalues Applications • Determine solutions to the matrix equation A x = λ x • Let’s put it in standard form, using λ x = λ I x : ( A − λ I ) x = 0 • B x = 0 has solutions other than x = 0 precisely when det ( B ) = 0. • hence we want det ( A − λ I ) = 0: Definition (Charachterisitc polynomial) The polynomial | A − λ I | is called the characteristic polynomial of A , and the equation | A − λ I | = 0 is called the characteristic equation of A . 5

Diagonalization Applications Example � 7 − 15 � A = 2 − 4 � 7 − 15 � � 1 0 � � 7 − λ � − 15 A − λ I = − λ = 2 − 4 0 1 2 − 4 − λ The characteristic polynomial is � � 7 − λ − 15 � � | A − λ I | = � � 2 − 4 − λ � � = ( 7 − λ )( − 4 − λ ) + 30 = λ 2 − 3 λ + 2 The characteristic equation is λ 2 − 3 λ + 2 = ( λ − 1 )( λ − 2 ) = 0 hence 1 and 2 are the only eigenvalues of A 6

Diagonalization Finding Eigenvectors Applications • Find non-trivial solution to ( A − λ I ) x = 0 corresponding to λ • zero vectors are not eigenvectors! Example � 7 − 15 � A = 2 − 4 Eigenvector for λ = 1: � 6 − 15 � � � � 5 � 1 − 5 RREF A − I = → · · · → 2 v = t , t ∈ R 2 − 5 0 0 2 Eigenvector for λ = 2: � 5 − 15 � � 1 − 3 � � 3 � RREF A − 2 I = → · · · → v = t , t ∈ R 2 − 6 0 0 1 7

Diagonalization Applications Example   4 0 4 A = 0 4 4   4 4 8 The characteristic equation is � � 4 − λ 0 4 � � � � | A − λ I | = 4 − λ 0 4 � � � � 4 4 8 − λ � � = ( 4 − λ )(( − 4 − λ )( 8 − λ ) − 16 ) + 4 ( − 4 ( 4 − λ )) = ( 4 − λ )(( − 4 − λ )( 8 − λ ) − 16 ) − 16 ( 4 − λ ) = ( 4 − λ )(( − 4 − λ )( 8 − λ ) − 16 − 16 ) = ( 4 − λ ) λ ( λ − 12 ) hence the eigenvalues are 4 , 0 , 12. Eigenvector for λ = 4, solve ( A − 4 I ) x = 0 :       4 − 4 0 4 1 1 0 − 1 RREF  →  , t ∈ R A − 4 I = 0 4 − 4 4 · · · → 0 0 1 v = t 1     4 4 8 − 4 0 0 0 0 8

Example   − 3 − 1 − 2 A = 1 − 1 1   1 1 0 The characteristic equation is � � − 3 − λ − 1 − 2 � � � � | A − λ I | = 1 − 1 − λ 1 � � � � 1 1 − λ � � = ( − 3 − λ )( λ 2 + λ − 1 ) + ( − λ − 1 ) − 2 ( 2 + λ ) = − ( λ 3 + 4 λ 2 + 5 λ + 2 ) if we discover that − 1 is a solution then ( λ + 1 ) is a factor of the polynomial: − ( λ + 1 )( a λ 2 + b λ + c ) from which we can find a = 1 , c = 2 , b = 3 and − ( λ + 1 )( λ + 2 )( λ + 1 ) = − ( λ + 1 ) 2 ( λ + 2 ) the eigenvalue − 1 has multiplicity 2

Diagonalization Eigenspaces Applications • The set of eigenvectors corresponding to the eigenvalue λ together with the zero vector 0 , is a subspace of R n . because it corresponds with null space N ( A − λ I ) Definition (Eigenspace) If A is an n × n matrix and λ is an eigenvalue of A , then the eigenspace of the eigenvalue λ is the nullspace N ( A − λ I ) of R n . • the set S = { x | A x = λ x } is always a subspace but only if λ is an eigenvalue then dim ( S ) ≥ 1. 10

Diagonalization Eigenvalues and the Matrix Applications Links between eigenvalues and properties of the matrix • let A be an n × n matrix, then the characteristic polynomial has degree n : p ( λ ) = | A − λ I | = ( − 1 ) n ( λ n + a n − 1 λ n − 1 + · · · + a 0 ) • in terms of eigenvalues λ 1 , λ 2 , . . . , λ n the characteristic polynomial is: p ( λ ) = | A − λ I | = ( − 1 ) n ( λ − λ 1 )( λ − λ 2 ) · · · ( λ − λ n ) Theorem The determinant of an n × n matrix A is equal to the product of its eigenvalues. Proof: if λ = 0 in the second point above, then p ( 0 ) = | A | = ( − 1 ) n ( − 1 ) n λ 1 λ 2 . . . λ n = λ 1 λ 2 . . . λ n 11

Diagonalization Diagonalization Applications Recall: Square matrices are similar if there is an invertible matrix P such that P − 1 AP = M . Definition (Diagonalizable matrix) The matrix A is diagonalizable if it is similar to a diagonal matrix; that is, if there is a diagonal matrix D and an invertible matrix P such that P − 1 AP = D Example � 7 − 15 � A = 2 − 4 How was such a matrix P found? � 5 3 � � − 1 � 3 P − 1 = P = 2 1 2 − 5 When is a matrix diagonalizable? � 1 0 � P − 1 AP = D = 0 2 13

Diagonalization General Method Applications • Let’s assume A is diagonalizable, then P − 1 AP = D where   0 · · · 0 λ 1 0 λ 2 · · · 0     D = diag ( λ 1 , λ 2 , . . . , λ n ) = ...   0 0 0   0 0 · · · λ n • AP = PD � � � � AP = A v 1 · · · v n = A v 1 · · · A v n   λ 1 0 · · · 0 0 λ 2 · · · 0   � � � �   PD = v 1 · · · v n  = λ 1 v 1 · · · λ n v n ...   0 0 0  0 0 · · · λ n • Hence: A v 1 = λ 1 v 1 , A v 2 = λ 2 v 2 , · · · A v n = λ n v n 14

Diagonalization Applications • since P − 1 exists then none of the above A v i = λ i v i has 0 as a solution or else P would have a zero column. • this is equivalent to λ i and v i are eigenvalues and eigenvectors and that they are linearly independent. • the converse is also true: suppose A has n lin. indep. eigenvectors and P be the matrix whose columns are the eigenvectors (then P is invertible) A v = λ v implies that AP = PD P − 1 AP = P − 1 PD = D Theorem An n × n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Theorem An n × n matrix A is diagonalizable if and only if there is a basis of R n consisting only of eigenvectors of A . 15

Diagonalization Applications Example � 7 − 15 � A = 2 − 4 and 1 and 2 are the eigenvalues with eigenvectors: � � � � 5 3 v 1 = v 2 = 2 1 � � 5 3 � � P = v 1 v 2 = 2 1 16

Diagonalization Applications Example   4 0 4 A = 0 4 4   4 4 8 has eigenvalues 4 , 0 , 12 and corresponding eigenvectors:       − 1 − 1 1  ,  , v 1 = 1 v 2 = − 1 v 3 = 1     0 1 2  − 1 − 1 1   4 0 0  P = 1 − 1 1 D = 0 0 0     0 1 2 0 0 12 We can choose any order, provided we are consistent:     − 1 − 1 1 0 0 0 P = − 1 1 1 D = 0 4 0     1 0 2 0 0 12 17

Diagonalization Geometrical Interpretation Applications • Let’s look at A as the matrix representing a linear transformation T = T A in standard coordinates, ie, T ( x ) = A x . • let’s assume A has a set of linearly independent vectors B = { v 1 , v 2 , . . . , v n } corresponding to the eigenvalues λ 1 , λ 2 , . . . , λ n , then B is a basis of R n . • what is the matrix representing T wrt the basis B ? A [ B , B ] = P − 1 AP � v 1 v 2 · · · v n � where P = (check earlier theorem today) • hence, the matrices A and A [ B , B ] are similar, they represent the same linear transformation: • A in the standard basis • A [ B , B ] in the basis B of eigenvectors of A � [ T ( v 1 )] B [ T ( v 2 )] B · · · [ T ( v n )] B � • A [ B , B ] = � for those vectors in particular T ( v i ) = A v i = λ i v i hence diagonal matrix � A [ B , B ] = D 18

Diagonalization Applications • What does this tell us about the linear transformation T A ?   b 1 b 2   For any x ∈ R n   [ x ] B = .  .  .   b n B its image in T is easy to calculate in B coordinates:       0 · · · 0 λ 1 b 1 λ 1 b 1 0 λ 2 · · · 0 b 2 λ 2 b 2             [ T ( x )] B = = . . ...       . . 0 0 0 . .       0 0 · · · λ n b n λ n b n B B • it is a stretch in the direction of the eigenvector v i by a factor λ i • the line x = t v i , t ∈ R is fixed by the linear transformation T in the sense that every point on the line is stretched to another point on the same line. 19