r r Prof. Inder K. Rana Room 112 B Department of Mathematics - PowerPoint PPT Presentation

▲✐♥❡❛r ❆❧❣❡❜r❛ Prof. Inder K. Rana Room 112 B Department of Mathematics IIT-Bombay, Mumbai-400076 (India) Email: ikr@math.iitb.ac.in Lecture 11 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Recall We showed that the matrix � − 5 � − 7 A = 2 4 has two eigenvalues λ = 2 , − 3 . Further we found eigenvectors � − 7 � � � 1 C 1 = and C 2 = . − 1 2 We defined � � 1 − 7 P := , − 1 2 checked that P is is invertible, and � 2 � 0 P − 1 A P = . − 3 0 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Diagonalizable This prompted us to ask the following: Question: Given a matrix A when does there exist an invertible matrix P such that P − 1 AP will be a diagonal matrix, and how to find P ? Before answering the following: Definition Let A be a n × n matrix with entries from I F = I R or I C . A is said to be diagonalizable if A if there exists an invertible matrix P F , such that P − 1 AP is a diagonal matrix. with eateries from I Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Diagonalizable The answer is given by the following: Theorem Let A be a n × n matrix. A is diagonalizable if and only if there exist scalars λ 1 , λ 2 , . . . , λ n ∈ I R and vectors R n C 1 , C 2 , . . . , C n ∈ I such that the following holds: (i) A C i = λ i C i ∀ 1 ≤ i ≤ n . Thus A has n-eigenvalues. (ii) The set { C 1 , . . . , C n } is linearly independent, and hence is a basis R n . of I Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Proof Since A is diagonalizable, there exists an invertible matrix P such that P − 1 AP = D , where D is a diagonal matrix. Let C 1 , . . . , C n be the columns of P . Then P = [ C 1 . . . C n ] . Since P is an invertible matrix, none of the column vectors C i = 0 . In fact, P = [ C 1 C 2 · · · C n ] being an invertible matrix , has rank n , and hence the set { C 1 , . . . , C n } is linearly independent. Let   λ 1 0 ...     D = .  ...      0 λ n Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Proof Then, P − 1 AP = D implies that AP = PD , A [ C 1 . . . C n ] = [ C 1 . . . C n ] D , i.e., i.e., [ A C 1 A C 2 . . . A C n ] = [ λ 1 C 1 λ 2 C 2 . . . λ n C n ] . Thus, A C i = λ i C i for all 1 ≤ i ≤ n . This proves one way. Conversely, F n such that { X 1 , . . . , X n } is a Let X 1 , X 2 . . . , X n be elements of I linearly independent set and for λ 1 , . . . , λ n ∈ I F , A X i = λ i X i , 1 ≤ i ≤ n . Define the matrix P as follows: P := [ X 1 X 2 · · · X n ] . Then rank ( P ) = n , and hence is an invertible matrix. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Proof cont... Further, AP = A [ X 1 · · · X n ] = [ A X 1 . . . A X n ] = [ λ 1 X 1 · · · λ n X n ] = PD , where   0 λ 1 ...     D := .   ...     0 λ n Hence, P − 1 AP = D , i.e., A is diagonalizable. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Diagonalization Problem: Given a n × n matrix A , when and how do find a linearly independent set of eigenvectors v 1 , v 2 , ..., v n for A ? Clearly, this set of n eigenvectors will form a basis. We start with a theorem about linear independence of eigenvectors. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Linear independence of eigenvectors Theorem Let λ 1 , λ 2 , ..., λ r be distinct eigenvalues of a square matrix A. Let v 1 , v 2 , ..., v r be a corresponding choice of eigenvectors. Then { v 1 , v 2 , ..., v r } is a linearly independent set. Proof: Let ℓ ≤ r be the smallest integer such that { v 1 , . . . , v ℓ } is linearly dependent. In that case { v 1 , . . . , v ℓ − 1 } is linearly independent, and hence there exists scalars α 1 ,. . . , α ℓ − 1 , not all zero, such that v ℓ = α 1 v 1 + . . . . + α ℓ − 1 v ℓ − 1 . (1) Then � ℓ − 1 ℓ − 1 ℓ − 1 � � � � λ ℓ v ℓ = A v ℓ = A α i v i = α i A v i = α i λ i v i (2) i = 1 i = 1 i = 1 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Linear Independence of eigenvectors Also, from (1) ℓ − 1 � λ ℓ v ℓ = λ ℓ α i v i . (3) i = 1 Thus, from (2) and (3) we have: ℓ − 1 � 0 = α i ( λ ℓ − λ i ) v i . i = 1 The linear independence of { v 1 , . . . , v ℓ − 1 } implies that α i ( λ i − λ ℓ ) = 0 , for all i . Since, α i is not zero for some i , we get λ i − λ ℓ = 0 for that i , which not possible as λ 1 , . . . , λ k are all distinct. Hence, { v 1 , . . . , v k } are linearly independent. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Diagonalization Theorem If a n × n matrix A , has n distinct eigenvalues then A is diagonalizable. If the characteristic polynomial has a multiple root the there could be a problem. � 1 � 1 Example: Let A = . 0 1 Let us find its eigenvalues and eigenvectors. Solution: The characteristic polynomial is D ( t ) = ( t − 1 ) 2 . Hence only one eigenvalue which is repeated. � 0 � 1 To find eigenvectors, we note that A − I 2 = A = 0 0 is itself in reduced row echelon form. � 1 � Further its null space is N ( A − I 2 ) = I , i.e., the span of the R 0 eigenvectors is only 1-dimensional, and hence there does not exist a R 2 consisting of eigenvectors. Thus A is not diagonalizable. basis of I Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Algebraic/geometric multiplicity, defect Henceforth, we denote N ( A − λ I n ) as E λ = E λ ( A ) and call it the λ -eigenspace of A . Definition (Algebraic multiplicity) Let λ be an eigenvalue of A . The multiplicity of λ as a root of the characteristic polynomial D A ( t ) is called the algebraic multiplicity of λ . We write this number as m λ = m λ ( A ) . Definition (Geometric multiplicity) Let λ be an eigenvalue of A . The dimension of the null space of A − λ I n is known as the geometric multiplicity of λ . We write this number as g λ = g λ ( A ) . Definition (Defect) Let λ be an eigenvalue of A . The difference m λ − g λ is known as the defect of λ and is denoted δ λ = δ λ ( A ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Diagonalization Theorem If the algebraic and the geometric multiplicities of an n × n matrix R n agree for every eigenvalue λ of A, then there exists a basis of I consisting of eigenvectors of A Proof: Let λ 1 , λ 2 , λ k be the distinct eigenvalues of A with multiplicities m 1 , m 2 , ..., m k respectively. Let B j = { v j 1 , v j 2 , ..., v j mj } be a basis of E λ j , j = 1 , 2 , ..., k . R n . Then B = � j B j will be an A -eigenbasis of I (Note that m 1 + m 2 + · · · + m k = deg D A ( t ) = n .) Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Similarity of matrices and diagonalization Definition (Similarity) Let A and B be two n × n matrices. We say A is similar to B if there is an invertible matrix P such that B = P − 1 AP . Theorem If A and B are similar, then both have the same eigenvalues with the same multiplicities -both algebraic and geometric. Theorem (Digonalization) Let A be diagonalizable i.e. each eigenvalue of A is defect-free. Then A is similar to a diagonal matrix whose diagonal entries are the eigenvalues of A, each occurring as many times as its multiplicity. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Example Example   1 2 − 2 − 4 Let A = 2 1  . Find the eigenvalues and a basis of  1 − 1 − 2 eigenvectors which diagonalizes A . Also write down a matrix X such that X − 1 AX is diagonal. Solution: D A ( λ ) = − λ ( λ 2 − 9 ) = ⇒ λ = 0 , ± 3 .  − 2    4 2 1 0 0  yields λ = − 3: Row reduction of A + 3 I = 2 4 − 4 0 1 − 1    1 − 1 1 0 0 0  0   as an eigenvector. v 1 = 1  1 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Example contd.  1 2 − 2  λ = 0: Row operations on A produces a matrix 0 1 0  .  0 0 0   2  is an eigenvector. Hence v 2 = 0  1  − 1    1 1 1  is an λ = 3: A − 3 I gives rise to 0 0 1  . Hence v 3 = 1   0 0 0 0 eigenvector.  0 2 1   . This enables us to write X = [ v 1 , v 2 , v 3 ] = 1 0 1  1 1 1     − 1 − 1 2 − 3 0 0 X − 1 =  . Finally, X − 1 AX =  . − 1 0 1 0 0 0   1 2 − 2 0 0 3 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Example Consider the matrix   3 0 0  . − 2 A = 4 2  − 2 1 5 The characteristic polynomial of A is � � 3 − λ 0 0 � � � � = − 2 4 − λ 2 � � � � − 2 5 − λ 1 � � = ( 3 − λ )[( 4 − λ )( 5 − λ ) − 2 ] ( 3 − λ )( 6 − λ )( λ − 3 ) . = Thus, A has two eigenvalues λ = 3 , 6 . Thus λ = 3 has algebraic multiplicity 2 and λ = 6 has algebraic multiplicity 1 . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Example Let us eigenvector for the eigenvalue λ = 3 . Note that  0 0 0   0 0 0   − 2 1 2   ∼  ∼  . A − 3 I − 2 − 2 = 1 2 1 2 0 0 0    − 2 1 2 − 2 1 2 0 0 0 Thus A − 3 I n has rank 1 and hence x 2     2 + x 3    x 2 , x 3 ∈ I N ( A − 3 I n ) = E 3 ( A ) = x 2 R  .  x 3  Prof. Inder K. Rana Department of Mathematics, IIT - Bombay