Diagonalization Marco Chiarandini Department of Mathematics & - PowerPoint PPT Presentation

DM554 Linear and Integer Programming Lecture 9 Diagonalization Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

Coordinate Change Diagonalization Outline Applications 1. More on Coordinate Change 2. Diagonalization 3. Applications 2

Coordinate Change Diagonalization Resume Applications • Linear transformations and proofs that a given mapping is linear • range and null space, and rank and nullity of a transformation, rank-nullity theorem • two-way relationship between matrices and linear transformations • change from standard to arbitrary basis • change of basis from B to B ′ 3

Coordinate Change Diagonalization Outline Applications 1. More on Coordinate Change 2. Diagonalization 3. Applications 4

Coordinate Change Diagonalization Change of Basis for a Lin. Transf. Applications We saw how to find A for a transformation T : R n → R m using standard basis in both R n and R m . Now: is there a matrix that represents T wrt two arbitrary bases B and B ′ ? Theorem Let T : R n → R m be a linear transformation and B = { v 1 , v 2 , . . . , v n } and B ′ = { v ′ n } be bases of R n and R m . 1 , v ′ 2 , . . . , v ′ Then for all x ∈ R n , [ T ( x )] B ′ = M [ x ] B where M = A [ B , B ′ ] is the m × n matrix with the ith column equal to [ T ( v i )] B ′ , the coordinate vector of T ( v i ) wrt the basis B ′ . Proof: x = P n × n ∀ x ∈ R n change B to standard [ x ] B B ↓ perform linear transformation T ( x ) = A x = AP n × n [ x ] B B in standard coordinates ↓ [ u ] B ′ = ( P m × m ) − 1 u ∀ u ∈ R m change to basis B ′ B ′ [ T ( x )] B ′ = ( P m × m ) − 1 AP n × n [ x ] B B ′ B M = ( P m × m ) − 1 AP n × n B ′ B 5

Coordinate Change Diagonalization Applications How is M done? • P B = [ v 1 v 2 . . . v n ] • AP B = A [ v 1 v 2 . . . v n ] = [ A v 1 A v 2 . . . A v n ] • A v i = T ( v i ) : AP B = [ T ( v 1 ) T ( v 2 ) . . . T ( v n )] • M = P − 1 B ′ AP B = P − 1 B ′ = [ P − 1 B ′ T ( v 1 ) P − 1 B ′ T ( v 2 ) . . . P − 1 B ′ T ( v n )] • M = [[ T ( v 1 )] B ′ [ T ( v 2 )] B ′ . . . [ T ( v n )] B ′ ] Hence, if we change the basis from the standard basis of R n and R m the matrix representation of T changes 6

Coordinate Change Diagonalization Similarity Applications Particular case m = n : Theorem Let T : R n → R n be a linear transformation and B = { x 1 , x 2 , . . . , x n } be a basis R n . Let A be the matrix corresponding to T in standard coordinates: T ( x ) = A x . Let � � P = x 1 x 2 · · · x n be the matrix whose columns are the vectors of B. Then for all x ∈ R n , [ T ( x )] B = P − 1 AP [ x ] B Or, the matrix A [ B , B ] = P − 1 AP performs the same linear transformation as the matrix A but expressed it in terms of the basis B . 7

Coordinate Change Diagonalization Similarity Applications Definition A square matrix C is similar (represent the same linear transformation) to the matrix A if there is an invertible matrix P such that C = P − 1 AP . Similarity defines an equivalence relation: • (reflexive) a matrix A is similar to itself • (symmetric) if C is similar to A , then A is similar to C C = P − 1 AP , A = Q − 1 CQ , Q = P − 1 • (transitive) if D is similar to C , and C to A , then D is similar to A 8

Example y y 1 1 x x − 2 2 − 2 2 − 1 − 1 • x 2 + y 2 = 1 circle in standard form • x 2 + 4 y 2 = 4 ellipse in standard form • 5 x 2 + 5 y 2 − 6 xy = 2 ??? Try rotating π/ 4 anticlockwise � 1 � � cos θ − sin θ � 2 − 1 √ √ 2 A T = = = P 1 1 sin θ cos θ √ √ 2 2 � 1 � � � � � 2 − 1 x X √ √ v = P [ v ] B ⇐ ⇒ = 2 1 1 y Y √ √ 2 2 X 2 + 4 Y 2 = 1

Coordinate Change Diagonalization Applications Example Let T : R 2 → R 2 : � x + 3 y �� x �� � T = y − x + 5 y What is its effect on the xy -plane? Let’s change the basis to �� 1 � � 3 �� B = { v 1 , v 2 } = , 1 1 Find the matrix of T in this basis: • C = P − 1 AP , A matrix of T in standard basis, P is transition matrix from B to standard � � � � � � � � C = P − 1 AP = 1 − 1 3 1 3 1 3 4 0 = 1 − 1 − 1 5 1 1 0 2 2 10

Coordinate Change Diagonalization Applications Example (cntd) • the B coordinates of the B basis vectors are � 1 � � 0 � [ v 1 ] B = , [ v 2 ] B = 0 1 B B • so in B coordinates T is a stretch in the direction v 1 by 4 and in dir. v 2 by 2: � � � � � � 4 0 1 4 [ T ( v 1 )] B = = = 4 [ v 1 ] B 0 2 0 0 B B • The effect of T is however the same no matter what basis, only the matrices change! So also in the standard coordinates we must have: A v 1 = 4 v 1 A v 2 = 2 v 2 11

Coordinate Change Diagonalization Resume Applications • Matrix representation of a transformation with respect to two given basis • Similarity of square matrices 12

Coordinate Change Diagonalization Outline Applications 1. More on Coordinate Change 2. Diagonalization 3. Applications 13

Coordinate Change Diagonalization Eigenvalues and Eigenvectors Applications (All matrices from now on 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 14

Coordinate Change 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 . 15

Coordinate Change 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 16

Coordinate Change 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: � � � � � � 1 − 5 6 − 15 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 17

Coordinate Change Diagonalization Applications Example   4 0 4 A = 0 4 4   4 4 8 The characteristic equation is � � 4 − λ 0 4 � � � � | A − λ I | = 0 4 − λ 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 18

Coordinate Change Diagonalization Applications Example  − 3 − 1 − 2  A = 1 − 1 1   1 1 0 The characteristic equation is � � − 3 − λ − 1 − 2 � � � � | A − λ I | = − 1 − λ 1 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 19

Coordinate Change 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. 20