5 Example: In a town, each year 30% married women get divorced20% single women get marriedIn the 1st year, 8000 married women 2000 single women.Total population remains constant, whererepresent the numbers of married & singlewomen after i years, respectively.
6 IfQuestion: Why does converge?Why does it converge to the same limitvector even when the initial condition isdifferent?
7 Ans: Choose a basisGiven an initial for somefor example,Question: How does one know choosing such a basis?
8 Def: Let . A scalar is said to be an eigenvalue or characteristic value of A ifsuch thatThe vector is said to be an eigenvectoror characteristic vector belonging to .is called an eigen pair of A.Question: Given A, How to compute eigenvalues& eigenvectors?
9 characteristic polynomial of A, of degree n in is an eigen pair ofNote that, is a polynomial, calledcharacteristic polynomial of A, of degree n inThus, by FTA, A has exactly n eigenvalues includingmultiplicities.is a eigenvector associated with eigenvalue while is eigenspace of A.
17 DiagonalizationGoal: Given find a nonsingular matrixS, such that is a diagonalmatrix.Question1: Are all matrices diagonalizable?Question2: What kinds of A are diagonalizable?Question3: How to find S if A is diagonalizable?
18 NOT all matrices are diagonalizable e.g., LetIf A is diagonalizable,nonsingular matrix S,
19 To answer Q2,Suppose that A is diagonalizable.nonsingular matrix S,LetThis gives a condition for diagonalizability and a way tofind S.
20 Theorem6.3.1: If are distinct eigenvalues of an matrix A with corresponding eigenvectors ,then are linearly independent.Pf: Suppose that are linearly dependentnot all zero,Suppose that
21 Theorem 6.3.2: Let is diagonalizable A has n linearly independent eigenvectors.Note : Similarity transformationChange of coordinatediagonalization
22 Remarks: Let , and(i) is an eigenpair of A for(ii) The diagonalizing matrix S is not unique becauseIts columns can be reordered or multiplied byan nonzero scalar(iii) If A has n distinct eigenvalues , A is diagonalizable.If the eigenvalues are not distinct , then may or maynot diagonalizable depending on whether A has nlinearly independent eigenvectors or not.(iv)
34 Def:(a) A is said to be Hermitian if(b) A is said to be skew-Hermitian if(c) A is said to be unitary if( i.e. its column vectors form an orthonormal set in )
35 (ii) Let and be two eigenpairs of A, Theorem6.4.1: Let , then(i)(ii) eigenvectors belonging to distinct eigenvaluesare orthogonal.Pf：(i) Let be an eigenpair of A,(ii) Let and be two eigenpairs of A,
36 (ii) Let and be two eigenpairs of A, Theorem6.4.1Pf：(ii) Let and be two eigenpairs of A,
37 Theorem: Let andthenPf：(i) Let be an eigenpair of A,
38 Theorem6.4.3 (Schur’s Theorem): Let , then unitary matrix , is upper triangular.Pf：The proof is by mathematical induction on n.(i) The result is obvious if n=1;(ii) Assume the hypothesis holds for k×k matrices;(iii) let A be a (k+1)×(k+1) matrix.
39 Proof of Schur’s Theorem Let be an eigenpair of A withUsing the Gram-Schmidt process, construct an orthonormalbasis ofLet
40 Proof of Schur’s Theorem By the induction hypothesis (ii)
41 Theorem6.4.4: (Spectral Theorem) If , then unitary matrix U that diagonalizes A .Pf：By Theorem ,unitary matrix ,where T is upper triangular .
42 Cor.6.4.5: Let A be real symmetric matrix . Then (i)(ii) an orthogonal matrix U,is a diagonal matrix.proof：
43 Example 4:Find an orthogonal matrix U that diagonalizes A.Sol : (i)(ii)
45 Note：If A has orthonormal eigenbasis Question:In addition to Hermitian matrices ,is there any other matrices possessingorthonormal eigenbasis?Note：If A has orthonormal eigenbasiswhere U is unitary &diagonal.
46 Def: A is said to be normal if Remark：Hermitian, Skew- Hermitian and Unitary matrices are all normal.
47 then ,where U is unitary & diagonal. Theorem6.4.6: A is normal A possesses anorthonormal eigenbasisPf：If A has an orthonormal eigenbasis,then ,where U is unitary & diagonal.
48 proof of Theorem6.4.6:By Th.6.4.3, unitary U,Compare the diagonal elements of