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 10 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Recall Definition (Orthonormal bases) R n . We say B Let B = { v 1 , v 2 , ..., v k } be a basis of a vector space V ⊆ I is an orthonormal basis of V , if B is an orthonormal set. Theorem R n . If the Let S = { v 1 , v 2 , ..., v k } be a set of non-zero vectors in I vectors are mutually orthogonal, then S is a linearly independent set. Theorem Let B = { u 1 , u 2 , ..., u k } be an orthonormal basis of a vector space V and x ∈ V. Then x = � x , u 1 � u 1 + � x , u 2 � u 2 + · · · + � x , u k � u k . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Constructing orthonormal basis How to construct orthonormal basis Suppose B = { v 1 , v 2 , ..., v k } is a a spanning set/basis of V . Step 1:Drop zero vectors, if any. Step 2: Define w 1 := v 1 . w 2 := v 2 − � v 2 , w 1 � � w 1 � 2 w 1 . Step 3: · · · having constructed { w 1 , ..., w j − 1 } , (dropping zero vectors if any) define j − 1 � v j , w ℓ � � w j = v j − � w ℓ � 2 w ℓ ℓ = 1 Step 4:Continue till all the vectors v 1 , v 2 , ..., v k have been used. Step 5:Normalize the vectors w 1 , w 2 , ..., w j obtained to get an orthonormal basis of V . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Bessel’s inequality Theorem Let { u 1 , u 2 , ..., u k } is an orthonormal set in V and v ∈ V Then k � � v , u ℓ � 2 ≤ � v , v � . ℓ Equality holds if and only if { u 1 , u 2 , ..., u k } be an orthonormal basis (called Parseval’s Identity). Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Isometries Definition Let V be an inner product space. A linear transformation T : V → V is called an isometry on V if � Tv , Tw � = � v , w � for all v , w ∈ V . That is, T is a linear map which preserves the angles between the vectors. Theorem Let V be an inner product space and T : V → V be a linear map. Then the following statements are equivalent: (i) T is an isometry. (ii) T preserves length, i.e., � Tv � = � v � for all v ∈ V . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Isometries Proof: Assume that T is an isometry. Then, for all u ∈ V , � Tu � 2 = � Tu , Tu � = � u , u � = � u � . Conversely, suppose that � Tv � = � v � ∀ v ∈ V . Then for all u , w ∈ V , � T ( u + w ) , T ( u + w ) � = � Tu + Tw , Tu + Tw � � Tu , Tu � + � Tw , Tu � + � Tu , Tw � + � Tw , Tw � = = � u , u � + 2 � Tu , Tw � + � w , w � . . . ( 1 ) . Also, since T is an isometry, � T ( u + w ) , T ( u + w ) � = � u + w , u + w � = � u , u � + 2 � u , w � + � w , w � . . . . ( 2 From (1) and (2) it follows that � Tu , Tw � = � u , w � ∀ u , w ∈ V . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Examples Example 1 Consider the map T : I R → I R , T ( x ) = α x , for x ∈ I R , where α ∈ I R is fixed. Then T is a linear map and � T ( x ) � = � x � , i.e., | α || x | = | x | if and only if | α | = 1 . Thus, T is an isometry if and only if | α | = 1 . Example 2 R 2 → I R 2 be defined by Let T : I � x � � � x cos θ + y sin θ T = y − x sin θ + y cos θ where 0 ≤ θ ≤ 2 π is fixed. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Examples Then, ( x cos θ + y sin θ ) 2 + ( − x sin θ + y cos θ ) 2 � Tx � 2 = x 2 ( cos 2 θ + sin 2 θ ) + y 2 ( sin 2 θ + cos 2 θ ) = x 2 + y 2 = � x � 2 . = R 2 . Hence, T is an isometry on I Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Matrix of an Isometry Theorem Let T : V → V be a linear map and B be an ordered orthonormal basis of V . Then, the following hold: (i) T is an isometry if and only if the column vectors of [ T ] B form an orthonormal set. (ii) T is an isometry if and only if row vectors of [ T ] B form an orthonormal set. Note: If C 1 , A = [ C 2 , ..., C n ] in the column form, then A t A = [ C 1 , C 2 , ..., C n ] t [ C 1 , C 2 , ..., C n ] = [ � C i , C j � ] . Definition A real matrix A = [ a ij ] n × n is said to be orthogonal if the column vectors of A form an orthonormal set in, A t A = I n . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Characterization of Orthogonal matrices Theorem Let A be a n × n real matrix. Then the following are equivalent: (i) A is orthogonal. (ii) A t A = I n . (iii) AA t = I n . Note: Thus, orthonormal matrices arise in matrix representation of isometries with respect to orthonormal basis. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

An example Consider the matrix: � − 5 � − 7 A = . 2 4 Then � − 5 � � � � � � � � � 1 − 7 1 2 1 A = = = 2 . − 1 − 1 − 2 − 1 2 4 � � 1 Thus the matrix A scales the vector . − 1 Question: Given a matrix A how to find all the vectors that are scaled by it? How to find all the scaling factors for a given matrix? Why it is important to answer these questions? Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Answers For the matrix A as above we saw � � � � 1 1 A = 2 . − 1 − 1 That is � � 1 ( A − 2 I ) = 0 − 1 R 2 | ( A − 2 I ) X = 0 } . Then V is a vector subspace of I R 2 Let V = { X ∈ I R 2 . of I Let us find its dimension: For that we have to find its nullity: � − 5 − 2 � − 7 A − 2 I = . 2 4 − 2 Clearly det( A − 2 I ) � = 0 , and hence A − 2 I is invertible. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Answers R 2 | ( A − 2 I ) X = 0 } has dimension 1 and is a basis Hence V = { X ∈ I of it. Question: Does there exist some other scalar λ and another nonzero R 2 such that ( A − λ I ) X = 0 ? vector X ∈ I Note that this implies that the matrix ( A − λ I ) is not invertible! Hence det ( A − λ I ) = 0 . That is � − 5 − λ � − 7 A − λ I = = 0 2 4 − λ � � 1 { } giving ( − 5 − λ )( 4 − λ ) + 14 − 7 = 0 − 1 Hence λ 2 + λ − 6 = 0 , ⇒ λ = 2 , − 3 . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Answers R 2 such that ( A + 3 I ) X = 0 . Let us find a vector X 0 ∈ I Since � − 5 + 3 � − 2 � � − 2 � � − 7 − 7 − 7 A + 3 I = = , 2 4 + 3 2 7 0 0 �� − 7 �� R 2 | ( A + 3 I ) X = 0 } = [ W := { X ∈ I . 2 The relations: � � � � 1 1 A = 2 − 1 − 1 and � − 7 � − 7 � � A = ( − 3 ) 2 2 can be written as Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Answers Let � − 7 � � � 1 C 1 = and C 2 = − 1 2 Then � − 5 � � 2 � � � � � − 7 1 − 7 1 − 7 0 A [ C 1 C 2 ] = = 2 4 − 1 2 − 1 2 0 − 3 � 2 � 0 = [ C 1 C 2 ] − 3 0 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Answers Noting that the matrix � � 1 − 7 P := − 1 2 is invertible, we have � 2 � 0 P − 1 A P = − 3 0 This prompts one 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 ? Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Eigenvalue problem This motivates our next definition Definition The scalar λ is called an eigenvalue of matrix A if there exists a 1 nonzero vector ] box such that A X = λ X . The vector X in the above is called an eigenvector of A 2 corresponding to the eigenvalue λ. Eigenvalue Problem ( EVP ): Given a matrix A , find its eigenvalues and eigenvectors corresponding to them Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Finding the eigenvalues Rewrite the EVP as ( A − λ I ) v = 0 , v � = 0 . By the theory of homogeneous linear equations, a solution exists ⇐ ⇒ the rank ρ ( A − λ I ) < n which happens ⇐ ⇒ the determinant | ( A − λ I ) | = 0 . Definition (Characteristic polynomial) The determinant � � a 11 − λ a 12 · · · a 1 n � � � � a 21 a 22 − λ · · · a 2 n � � � � D A ( λ ) = D ( λ ) = . . . � � . . . . . . � � � � · · · a nn − λ a n 1 a n 2 � � which is a polynomial of degree n is called thecharacteristic polynomial of the matrix A . The highest degree coefficient is ( − 1 ) n . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay

Characteristic roots/eigenvalues Definition (Characteristic roots/eigenvalues) The possibly complex roots of the characteristic polynomial D A ( λ ) are called the characteristic roots or the eigenvalues of A . There are n such roots when counted with multiplicity. Example 1: Find the positive eigenvalues and corresponding eigenvectors for   0 9 5  . A = 0 . 4 0 0  0 0 . 4 0 Solution: pause − D ( λ ) = λ 3 − 18 5 λ − 4 5 = ⇒ λ = 2 is a root. The other roots are from − D ( λ ) ( λ − 2 ) = λ 2 + 2 λ + 2 5 = 0 which are negative and hence discarded. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay