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Chapter 3 Linear Algebra February 29 Matrices 3.2 Matrices; Row reduction Standard form of a set of linear equations: 2 x 5 y z 2 x y 2z 1 x 5z 3 Matrix of coefficients: 2 5 1 M 1 1 2 1 0 5 Augmented matrix: 2 5 1 2 A 1 1 2 1 1 0 5 3 Elementary row operations: 1) Row switching. 2) Row multiplication by a nonzero number. 3) Adding a multiple of a row to another row. These operations are reversible. 1 Solving a set of linear equations by row reduction: * * * * * Method : As close as possible * * * * 0 * * * * 0 1 0 0 * * * * * * * 0 * * * 0 * * * 1 1 * * 0 0 1 * 0 * * * * * * Row reduced matrix 0 * * * * * 1 0 0 * 1 0 * 0 1 0 * 0 1 * 0 0 1 * Example : 2 1 2 1 2 5 1 2 1 1 2 1 (2)(1)2 1 1 1 1 (1)(2) (3)(1) (2)/3 0 3 3 0 0 1 1 0 1 1 2 1 2 5 1 2 1 0 5 3 1 0 5 3 0 1 3 2 0 1 3 2 1 1 2 1 1 1 2 1 1 1 2 1 (2) ( 3) (3) 0 1 1 0 (3)/2 0 1 1 0 (2) 0 1 0 1 0 0 2 2 0 0 1 1 0 0 1 1 1 0 0 2 x 2 3 )2 (1) (2) ( 0 1 0 1 y 1 0 0 1 1 z 1 2 Rank of a matrix: The number of the nonzero rows in the row reduced matrix. It is the maximal number of linearly independent row (or column) vectors of the matrix. Rank of A = Rank of AT (where (AT)ij= (A)ji, transpose matrix). Possible cases of the solution of a set of linear equation: Solving m equations with n unknowns: 1) If Rank M < Rank A, the equations are inconsistent and there is no solution. * * * * 0 * * * 0 0 0 * 2) If Rank M = Rank A = n, there is one solution. * * * * 0 * * * 0 0 * * 3) If Rank M = Rank A = R<n, then R unknowns can be expressed by the remaining n−R unknowns. * * * * 0 * * * 0 0 0 0 3 Read: Chapter 3: 1-2 Homework: 3.2.4,8,10,12,15. Due: March 11 4 March 7 Determinants 3.3 Determinates; Cramer’s rule Determinate of an n×n matrix: a11 a12 * a1n a11 a12 * a1n A * * * * , detA | A | * * * * a an1 * * ann n1 * * ann Minor: a11 a12 * a1n A * * * * , an1 * * ann j a11 a12 * a1n M ij * * * * i an1 * * ann Cofactor: Cij (1)i j M ij Determinate of a 1×1 matrix: a a 1 j Definition of the determinate of an n×n matrix: A a1 j C1 j (1) a1 j M 1 j j j 5 Equivalent methods: 1) A determinate can be expanded by any row or any column: A aijCij aijCij j i 2) |A|=|AT|. Example p90.1 Triple product of 3 vectors: A1 A2 A3 A ( B C) C ( A B) B (C A ) B1 B2 B3 C1 C2 C3 (signed ) volume of the parallelop iped formed by the 3 vectors. Useful properties of determinants: 1) A common factor in a row (column) may be factored out. 2) Interchanging two rows (columns) changes the sign of the determinant. 3) A multiple of one row (column) can be added to another row (column) without changing the determinant. 4) The determinate is zero if two rows (columns) are identical or proportional. Example p91.2. 6 Cramer’s rule in solving a set of linear equations: a11 x1 a12 x2 a13 x3 c1 a11 a12 a21 x1 a22 x2 a23 x3 c2 x1 a21 a22 a31 a32 a31 x1 a32 x2 a33 x3 c3 c1 a12 a13 c1 a12 a13 a11 x1 a12 x2 a13 x3 a12 a13 a23 a21 x1 a22 a23 a21 x1 a22 x2 a23 x3 a22 a23 a33 a33 a31 x1 a32 x2 a33 x3 a32 a33 a13 a11 a11 x1 a12 a31 x1 a12 a32 a13 a13 c2 a22 a23 x1 c2 a22 a23 a21 a22 a23 , and similar solutions for x2 and x3 . c3 a32 a33 a32 a33 a31 a32 a33 c3 Theorem: For homogeneous linear equations, the determinant of the coefficient matrix must be zero for a nontrivial solution to exist. a11 a12 a13 0 a12 a13 a11x1 a12 x2 a13 x3 0 a21x1 a22 x2 a23 x3 0 If a21 a22 a23 0, x1 0 a22 a23 a31x1 a32 x2 a33 x3 0 a31 a32 a33 0 a32 a33 a11 a12 a13 a21 a22 a23 0. a31 a32 a33 7 Read: Chapter 3: 3 Homework: 3.3.1,10,15,17. (No computer work is needed.) Due: March 25 8 March 11 Vectors 3.4 Vectors Vector: A quantity that has both a magnitude and a direction. Geometrical representation of a vector: An arrow with a length and a direction . Addition and subtraction: Vector addition is commutative and associative. Algebraic representation of a vector: A ( Ax , Ay , Az ) Ax ˆi Ay ˆj Az kˆ A B ( Ax Bx )ˆi ( Ay By )ˆj ( Az Bz )kˆ cA cAx ˆi cAy ˆj cAz kˆ Magnitude of a vector: A A Ax2 Ay2 Az2 Note: A is a vector and A is its length. They should be distinguished. 9 Scalar or dot product: A B AB cos Ax Bx Ay By Az Bz B (Proof) Vector or cross product: A B AB sin , with its direction A determined by the right hand rule. Cross product in determinant form: ˆi ˆj kˆ A B Ax Ay Az (Proof) Bx By Bz A×B B A Parallel and perpendicular vectors: Ax Ay Az A // B A B 0. Bx By Bz A B Ax Bx Ay By Az Bz 0 A B 0. Relations between the basis vectors: ˆi ˆi 1, ˆi ˆj 0, ˆi ˆi 0, ˆi ˆj kˆ . Examples p102.3, p105.4. Problems 4.5,26. 10 Read: Chapter 3: 4 Homework: 3.4.5,7,12,18,26. Due: March 25 11 March 21 Lines and planes 3.5 Lines and planes z Equations for a straight line: r r0 At x x0 at y y0 bt z z ct 0 x x0 y y0 z z0 a b c Equations for a plane: N (r r0 ) 0 a( x x0 ) b( y y0 ) c( z z0 ) 0 ax by cz d ( ax0 by0 cz0 ) r−r0 ( x, y , z ) ( x0 , y0 , z0 ) A y x N r−r0 ( x, y , z ) ( x0 , y0 , z0 ) Examples p109.1, 2. 12 Distance from a point to a plane: N PR PQ cos PQ PQ n N P n Q Example p110.3. R Distance from a point to a line: A PR PQ sin PQ PQ u A P A Example p110.4. Q Distance between two skew lines: Let FG be the shortest distance, then it must be perpendicular to both lines (proof). FG PQ n PQ P F AB AB Example p110.5,6. Problems 5.12,18,42. R A n Q G B 13 Read: Chapter 3: 5 Homework: 3.5.7,12,18,20,26,32,37,42. Due: April 1 14 March 23,25 Matrix operations 3.6 Matrix operations Matrix equation: a b 2 3 a 2, b 3, c 4, d 5. c d 4 5 Multiplication by a number: a b ka kb k c d kc kd while k a b ka kb ka b . c d c d kc d Matrix addition: a b e c d g f ae b f h c g d h 15 Matrix multiplication: ABij Aik Bkj . k Note: 1) The element on the ith row and jth column of AB is equal to the dot product between the ith row of A and the jth column of B. 2) The number of columns in A is equal to the number of rows in B. Example: a b e c d g f ae bg af bh h ce dg cf dh More about matrix multiplication: 1) The product is associative: A(BC)=(AB)C 2) The product is distributive: A(B+C)=AB+AC 3) In general the product is not commutative: ABBA. [A,B]=AB−BA is called the commutator. Unit matrix: 1 0 I 0 1 Zero matrix: 0 0 0 0 0 Product theorem: det( AB) det A det B 16 Solving a set of linear equations by matrix operations: 2 x 5 y z 2 2 5 1 x 2 x y 2 z 1 1 1 2 y 1 x 5 z 3 1 0 5 z 3 M r k r M 1 k. Matrix inversion: 1) M−1 is the inverse of M if MM−1= M−1M =I=1. 2) Only square matrix can be inversed. 3) det (M M−1) = det(M) det( M−1) = det(I)=1, so det (M)0 is necessary for M to be invertible. CT Calculating the inverse matrix: M . det M Proof : det M, if i j. (MC T )ij M ik CT kj M ik C jk 0, if i j. (p3.8) k k CT T 1 MC I det M M . det M 1 Example p120.3. 17 Three equivalent ways of solving a set of linear equations: 1) Row reduction 2) Cramer’s rule 3) Inverse matrix: r=M-1k Equivalence between r=M-1k and the Cramer’s rule: a11 a12 a21 a22 a 31 a32 a13 x k1 CT 1 a23 y k 2 M r k r M k k det M a33 z k3 k1 a12 a13 k 2 a22 a23 T T T k a32 a33 C11 k1 C12 k 2 C13 k3 k1C11 k 2 C 21 k3C31 x 3 . a11 a12 a13 det M det M a21 a22 a31 a32 a23 a33 Cramer’s rule is an actual realization of r=M-1k. 18 Gauss-Jordan method of matrix inversion: Let (MLp MLp-1 MLp-2 … ML2 ML1 ) M = MLM = I be the result of a series of elementary row operations on M, then (MLp MLp-1 MLp-2 … ML2 ML1 ) I = MLI = M−1. That is, M1 M I I M1 . 1 2 1 0 ( 2 )(1)3 1 2 1 0 (1)( 2 ) 1 0 2 1 Example : 0 2 3 1 3 4 0 1 0 2 3 1 1 1 0 2 . 0 1 3 / 2 1/ 2 1 1 0 1 2 2 Test : . 3 4 3 / 2 1 / 2 0 1 ( 2 )( 1 / 2 ) Equivalence between row reduction and r=M-1k: (M Lp M Lp1 M Lp2 M L2 M L1 )M M 1 M I (M Lp M Lp1 M Lp2 M L2 M L1 )( M, k ) M 1 (M, k ) (I, r ). Row reduction is to decompose the M−1 in r=M−1k into many steps. 19 Rotation matrices: Rotation of vectors X cos Y sin sin x , cos y y ( X ,Y ) ( x, y ) cos sin x . or R M r with M sin cos cos 2 sin 2 cos 1 sin 1 x cos(1 2 ) sin( 1 2 ) x . M 2 M1r sin 2 cos 2 sin 1 cos 1 y sin( 1 2 ) cos(1 2 ) y Functions of matrices: Expansion or power series expansion is implied. Examples: (A B) 2 A 2 AB BA B2 k 2 A 2 k 3A 3 exp( kA) 1 kA 2! 3! 1 1 A A2 1 A 20 Read: Chapter 3: 6 Homework: 3.6.9,13,15,18,21 Due: April 1 21 March 28, 30 Linear operators 3.7 Linear combinations, linear functions, linear operators Linear combination: aA + bB Linear function f (r): f (r1 r2 ) f (r1 ) f (r2 ), Examples : f (r ) A r, F(r ) br f (ar ) a f (r ) Linear operator O: O( f g ) O( f ) O( g ), O(af ) a O( f ) Here f and g can be numbers, functions, vectors, matrices, etc. d Example : f dx Example p125.1; Problem 7.15 Linear transformation: X a b x or R M r. M(r1 r2 ) Mr1 Mr2 , M( kr) k (Mr). Y c d y The matrix M is a linear operator representing a linear transformation. 22 Orthogonal transformation: An orthogonal transformation preserves the length of a vector. Orthogonal matrix : The matrix for an orthogonal transformation is an orthogonal matrix. Theorem: M is an orthogonal matrix if and only if MT = M−1. Proof : (Mr) T (Mr) r T r r T M T Mr r T r M T M I M T M 1. Theorem: det M=1 if M is orthogonal. Proof : det I det( MM 1 ) det( MM T ) det(M)det( M T ) [det(M) ]2 1 det(M) 1. 23 2×2 orthogonal matrix: 2 a b cos sin 2 a b 1 c d sin cos a b a c 1 c 2 d 2 1 c d b d ac bd 0 sin cos sin cos 0 sin( ) 0 a b cos sin 1) for , or det M 1. c d sin cos cos sin r cos r cos( ) . M is a rotaion. sin cos r sin r sin( ) change the sign of cos sin a b cos sin 2) for , or det M 1 . c d sin cos sin cos cos sin r cos r cos( ) . M is a reflection with respect to the line. 2 sin cos r sin r sin( ) cos Espe cially sin sin r cos( / 2) r cos( / 2) . cos r sin( / 2) r sin( / 2) Conclusion: A 2×2 orthogonal matrix corresponds to either a rotation (with det M=1) or a reflection (with det M=−1). 24 Two-dimensional rotation: Rotating the vector : X cos (Active transform ation :) Y sin Rotating the axes : (Change of basis :) x' y' y sin x cos y cos sin sin x cos y Two-dimensional reflection: x' y' cos sin ( X ,Y ) x y' y ( x, y ) sin x is a reflection cos y with respect to the ( x, y ) x' line, or y x tan . 2 2 y ( x' , y ' ) Example p128.3. 1 Note that A 2 3 2 x ( x, y ) 4 3 cos 3 2 4 1 sin 3 2 4 sin 3 4 cos 3 /2 x 25 Read: Chapter 3: 7 Homework: 3.7.9,15,22,26. Due: April 8 26 April 1 Linear dependence and independence 3.8 Linear dependence and independence Linear dependence of vectors: A set of vectors are linearly dependent if some linear combination of them is zero, with not all the coefficients equal to zero. 1. If a set of vectors are linearly dependent, then at least one of the vectors can be written as a linear combination of others. b c aA bB cC 0, a 0 A B C a a 2. If a set of vectors are linearly dependent, then at least one row in the row-reduced matrix of these vectors equals to zero. The rank of the matrix is then less than the number of rows. b c aA bB cC 0, a 0 A B C 0. a a Example: Any three vectors in the x-y plane are linearly dependent. E.g. (1,2), (3,4), (5,6). Linear independence of vectors: A set of vectors are linearly independent if any linear combination of them is not zero, with not all the coefficients equal to zero. 27 Linear dependence of functions: A set of functions are linearly dependent if some linear combination of them is always zero, with not all the coefficients equal to zero. Examples: sin x and cos x are linearly independen t. 1, x and x 2 are linearly independen t. sin 2 x, cos 2 x and 1 are linearly dependent. Theorem: If the Wronskian of a set of functions f1 ( x ) f ' ( x) W 1 ( n 1) f1 ( x ) f2 ( x) f2 ' ( x) ( n 1) f2 ( x) fn ( x) f 'n ( x ) 0, ( n 1) fn ( x) then the functions are linearly independent . 28 Proof : If f1 ( x ), f 2 ( x ), , f n ( x ) are linearly dependent, then there exit ki (supposed to be k1 ) 0 and n k j 1 j n n j 1 j 1 f j ( x ) 0. Now k j f j ' ( x ) 0, , k j f j ( x ) 0. 0 0 f2 ( x) f2 ' ( x) ( n 1) 0 f2 ( x) k1 f1 ( x ) f2 ( x) f1 ' ( x ) f2 ' ( x) ( n 1) ( n 1) f1 ( x ) f 2 ( x ) fn ( x) f 'n ( x ) (n) f1 ( x ) ( n 1) f1 ' ( x ) fn ( x) 0 0 W fn ( x) W (n) f 'n ( x ) f1 ( x ) ( n 1) fn ( x) f2 ( x) f2 ' ( x) (n) f2 ( x) fn ( x) f 'n ( x ) 0. (n) fn ( x) Examples p133.1,2. Note: W=0 does not always imply the functions are linearly dependent. E.g., x2 and x|x| about x=0. However, when the functions are analytic (infinitely differentiable), which we meet often, W=0 implies linear dependence. 29 Homogeneous equations: * * * 0 * 0 0 0 * * * 0 row reduction 0 * 0 0 or 0 * * 0 . * * * 0 * * * 0 0 0 * 0 0 0 0 0 1. Homogeneous equations always have the trivial solution (all unknowns=0). 2. If Rank M=Number of unknowns, the trivial solution is the only solution. 3. If Rank M< Number of unknowns, there are infinitely many solutions. Theorem: A set of n homogeneous equations with n unknowns has nontrivial solutions if and only if the determinant of the coefficients is zero. Proof: 1. det M ≠0 r =M-10=0. 2. Only trivial solution exists Columns of M are linearly independent det M ≠0. Examples p135.4. 30 Read: Chapter 3: 8 Homework: 3.8.7,10,13,17,24. Due: April 8 31 April 4 Special matrices 3.9 Special matrices and formulas Transpose matrix AT of A: AT ij A ji Complex conjugate matrix A* of A: A* ij Aij * Adjoint (transpose conjugate) matrix A+ of A: A (A T )* (A* )T , A ij A* ji A ji * Inverse matrix A-1 of A: A-1A= AA-1=1. Symmetric matrix: A=AT (A is real) Orthogonal matrix: A−1 =AT (A is real) Hermitian matrix: A =A+ Unitary matrix: A-1 =A+ Normal matrix: AA+=A+A, or [A,A+]=0. 32 Index notation for matrix multiplication: ABij A ik Bkj k 1, if i j. 0, if i j. Kronecker d symbol: d ij Exercises on index notations: Associative law for matrix multiplication: A(BC)=(AB)C Proof : ABC ij Aik BC kj Aik Bkl Clj ABil Clj ABCij k k ,l l Transpose of a product: (AB)T=BTAT Proof : ABijT AB ji A jk Bki ATkj BikT BikT ATkj BT AT ij k k k Corollary: (ABC)T=CTBTAT 33 Inverse of a product: (AB)-1=B-1A-1 Proof : (AB) (B-1A-1 ) A(BB -1 )A -1 AA -1 1 (AB) -1 B-1A-1. Corollary: (ABC)-1=C-1B-1A-1 Trace of a matrix: Tr A Aii i Trace of a product: Tr(AB)=Tr(BA) Proof : Tr(AB) ( AB)ii Aij B ji B jiAij Tr(BA) i i, j i, j Corollary: Tr(ABC)=Tr(BCA)=Tr(CAB) 34 Read: Chapter 3: 9 Homework: 3.9.2,4,5,23,24. Due: April 15 35 April 6 Linear vector spaces 3.10 Linear vector spaces n-dimensional vectors: A ( A1 , A2 ,, An ) Linear vector space: Several vectors and the linear combinations of them form a space. Subspace: A plane is a subspace of 3-dimensional space. Span: A set of vectors spans the vector space if any vector in the space can be written as a linear combination of the spanning set. Basis: A set of linearly independent vectors that spans a vector space. Dimension of a vector space: The number of basis vectors that span the vector space. Examples p143.1; p144.2. 36 n Inner product of two n-dimensional vectors: A B Ai Bi i 1 Length of an n-dimensional vector: A A A n A 2 i i 1 n Two n-dimensional vectors are orthogonal if A B Ai Bi 0. i 1 Schwarz inequality: A B AB. That is n AB i 1 i i n n A B i 1 2 i i 1 2 i . Proof : A B , e 2 , which are unit vecto rs along A and B. A B To prove A B AB, we then need to prove e1 e 2 1. Suppose A 0, B 0. Let e1 Construct C e1 (e1 e 2 )e 2 , which is the component of e1 othogonal to e 2 . Then C C 1 2(e1 e 2 ) 2 (e1 e 2 ) 2 1 (e1 e 2 ) 2 C 2 0 e1 e 2 1. The equal sign holds only when A// B. 37 Orthonormal basis: A set of vectors form an orthonormal basis if 1) they are mutually orthogonal and 2) each vector is normalized. Gram-Schmidt orthonormalization: Starting from n linearly independent vectors v1, v 2 , ,v n , we can construct an orthonormal basis set w1, w 2 , ,w n . v w1 1 , v1 v v 2 w1 w1 w2 2 , v 2 v 2 w1 w1 v v 3 w1 w1 v 3 w 2 w 2 w3 3 , v 3 v 3 w1 w1 v 3 w 2 w 2 k 1 wk v k v k w i w i i 1 k 1 v k v k w i w i . i 1 Example p146.4. 38 Bra-ket notation of vectors: A1 * * A A A2 , A A ( A1 A2 ). n A B A B Ai* Bi i 1 Complex Euclidean space: n Inner product: A B Ai* Bi i 1 Length: A n A A AA i 1 * i i n Orthogonal vectors: A B Ai* Bi 0 i 1 n Schwarz inequality: A B i 1 * i i n n A A B B i 1 * i i i 1 * i i Example p146.5. 39 Read: Chapter 3: 10 Homework: 3.10.1,10. Due: April 15 40 April 8, 11 Eigenvalues and eigenvectors 3.11 Eigenvalues and eigenvectors; Diagonalizing matrices Eigenvalues and eigenvectors: For a matrix M, if there is a nonzero vector r and a scalar l such that Mr lr, then r is called an eigenvector of M, and l is called the corresponding eigenvalue. M only changes the “length” of its eigenvector r by a factor of eigenvalue l, without affecting its “direction”. Mr lr (M l )r 0. For nontrivial solutions of this homogeneous equation, we need M 11 l M 21 det( M l ) M n1 M 12 M 22 l M n2 M 1n M 2n 0. M nn l This is called the secular equation, or characteristic equation. 41 4 3 . Example: Calculate the eigenvalues and eigenvectors of M 2 5 4l 3 det( M l ) 0 l1 2, l2 7. 2 5l 4 l1 3 x 3 0 2 x 3 y 0 r1 2 2 5 l1 y 3 x 4 l2 1 0 x y 0 r2 1 2 5 l2 y 5 2 . Example: Calculate the eigenvalues and eigenvectors of M 2 2 5l 2 det( M l ) 0 l1 1, l2 6. 2 2l 5 l1 2 x 1 0 2 x y 0 r1 2 2 2 l1 y 2 x 5 l2 2 0 x 2 y 0 r2 1 2 2 l2 y 42 Similarity transformation: Let operator M actively change (rotate, stretch, etc.) a vector. The matrix representation of the operator depends on the choice of basis vectors: In the old basis : R Mr; In the new basis : R ' M' r '. y' y r (r') M (M') Let matrix C change the basis (coordinate transformation): R (R') r' Cr ; R ' CR. x' Question: M' f (M, C) ? x R' M' r' M' Cr 1 M' C CM M' CMC . R' CR CMr M′ =CMC-1 is called a similar transformation of M. M′ and M are called similar matrices. They are the same operator represented in different bases that are related by the transformation matrix C. That is: If r′ =Cr, then M′ = CMC-1. Theorem: A similarity transformation does not change the determinate or trace of a matrix: det M' det M Tr M' Tr M r C M' M R r' C R' 43 Diagonalization of a matrix: For a 2 2 matrix M, suppose Mr1 l1r1 and Mr2 l2r2 , then x2 l1 x1 l2 x2 x1 x2 l1 0 y2 l1 y1 l2 y2 y1 y2 0 l2 x x2 l 0 (columns of eigenvecto rs), D 1 (diagnal of eigenvalue s), then Let C 1 y y 0 l 2 2 1 MC CD C 1MC D. x M 1 y1 Theorem: A matrix M may be diagonalized by a similarity transformation C-1MC=D , where C consists of the column eigenvectors of M, and the diagonalized matrix D consists of the corresponding eigenvalues. That is, the diagonization equation C-1MC=D just summarizes the eigenvalues and eigenvectors of M. An n n matrix M can be diagonized by C 1MC D, where Mri li ri , and r r rn , C 1 2 l1 0 0 l2 D 0 0 0 0 0 . 0 ln 0 44 4 3 3 1 . l1 2, r1 ; l2 7, r2 . Example 1 : M 2 5 2 1 3 1 1 1 1 1 2 0 , C , D . C 2 1 2 3 0 7 5 1 1 1 4 3 3 1 2 0 . C MC D 5 2 3 2 5 2 1 0 7 1 5 2 1 1 1 2 . l1 1, r1 ; l2 6, r2 . Example 2 : M 2 2 2 5 5 1 1 0 1 1 2 1 1 1 2 , C , D . C 5 2 1 5 2 1 0 6 C 1MC D 1 1 2 5 2 1 1 2 1 0 . 2 1 2 2 2 1 0 6 5 5 More examples : L I ω, D E. 45 More about the diagonalization of a matrix C-1MC=D: (2×2 matrix as an example): 1. D describes in the (x', y') system the same operation as M describes in the (x, y) system. 2. The new x', y' axes are along the eigenvectors of M. 1 x 1 x1 -1 x Proof : e x ' C C 0 y y 0 y1 x2 1 x1 y2 0 y1 3. The operation is more clear in the new system: x' l 0 x' l1 x' D 1 y' 0 l2 y' l2 y' 46 Diagonalization of Hermitian matrices: 1. The eigenvalues of a Hermitian matrix are always real. 2. The eigenvectors corresponding to different eigenvalues of a Hermitian matrix are orthogonal. Proof : H i li i j H i li j i If i j then li l*i . * (li l j ) j i 0 * H H j H i Hj i l j j i If i j then i j 0. 3. {A matrix has real eigenvalues and can be diagonalized by a unitary similarity transformation}if and only if {it is Hermitian}. Proof : 1) Suppose U 1MU D, with U 1 U and diagonal D D*. U 1MU D ( U 1MU) U M (U 1 ) U 1M U D D M UDU -1 M. 2) Suppose H H , Hri liri , then li li , r j ri rjk* rik d ij . * k Construct matrices C and D so that Cij rji , Dij lid ij , then C1HC D, and D D*. r r d ij Ckj* Cki C jkCki CC 1 C C1. * jk ik k k k Example p155.2. 47 Corollary: {A matrix has real eigenvalues and can be diagonalized by an orthogonal similarity transformation} if and only if {it is symmetric}. Proof : 1) Suppose O 1MO D, with O1 OT and diagonal D D*. O1MO D (O 1MO)T OT M T (O 1 )T O 1M T O DT D M T ODO-1 M. * 2) Suppose S ST , Sri liri , then li li , r j ri rjk rik d ij . k Construct matrices C and D so that Cij rji , Dij lid ij , then C1SC D, and D D*. rjk rik dij CkjCki C TjkCki CTC 1 CT C1. k k k 5 2 1 1 1 2 . l1 1, r1 ; l2 6, r2 . Example : M 2 2 2 5 5 1 1 0 1 1 2 1 1 1 2 . C ,C , D 5 2 1 5 2 1 0 6 1 1 2 5 2 1 1 2 1 0 . C 1MC D 2 1 2 2 2 1 0 6 5 5 li l*i Notice that M M 1 . T C C T 48 Read: Chapter 3: 11 Homework: 3.11.3,13,14,19,32,33,42. Due: April 22 49