Edwards-Penney Chapter 6, sections 6.1, 6.2 Topics, Definitions, Theorems, Examples 6.1 Introduction to Eigenvalues ==== DEF. A pair (lambda,v) with vector v nonzero and lambda a scalar is called an Eigenpair for matrix A provided Av=lambda v. Then lambda is called and eigenvalue and v a corresponding eigenvector. REMARKS. Two eigenpairs (lambda,v1) and (lambda,v2) are considered to be the same if v1 is a scalar multiple of v2. The set of all eigenpairs with lambda fixed is a subspace. EXAMPLE. Verify that v1=<2,1> and lambda1=2 make an eigenpair (lambda1,v1) for the matrix A:=Matrix([[5,-6],[2,-2]]). Repeat for v2=<3,2>, lambda2=1 for a second eigenpair (lambda2,v2). DETAILS. In each case, multiply out Av and compare to lambda v. FRACTION-FREE If v=<3/2,5/3> is an eigenvector for lambda, then so is (6)v=<9,10>. We would report the eigenpair as (lambda,<9,10>), a fraction-free answer. Maple, mathematica and matlab will report eigenvectors in fraction free form. GEOMETRY There is a geometrical meaning to an eigenvector. If matrix A is 2x2 and represents an ellipse equation, then the eigenvalues are physical invariants connected to the semiaxis lengths and the eigenvectors are the semiaxis directions. CHARACTERISTIC EQUATION An eigenpair is (lambda,v) means v is nonzero and (A-lambda I)v=0. The latter has a nonzero solution v only for the infinite solution case which is equivalent to determinant zero. Therefore: THEOREM 1. For an eigenpair, lambda is determined from the CHARACTERISTIC EQUATION |A - lambda I|=0. THEOREM. The characteristic polynomial |A - lambda I| is by cofactor expansion a polynomial in variable lambda of degree n, when A is an nxn matrix. The characteristic equation has exactly n roots counting multiplicities, which accounts for all possible eigenvalues of A. The form of the characteristic equation is (-lambda)^n + c1(-lambda)^(n-1) + ... + cn (1) = 0 ALGORITHM for EIGENPAIRS of MATRIX A 1. Solve for all values lambda in the polynomial equation |A - lambda I|=0, the eigenvalues of A. 2. For each distinct value lambda found in item 1, solve for all possible nonzero vectors v in the homogeneous equation (A - lambda I)v=0 3. Report all eigenpairs (lambda,v) found in items 1,2. EXAMPLE. Find all eigenpairs for A:=Matrix([[5,7],[-2,-4]]). ANSWER: (-2,<1,-1>) and (3,<7,-2>) EXAMPLE. Find all eigenpairs for A:=Matrix([[0,8],[-2,0]]). ANSWER: (4i,<2i,-1>) and (-4i,<2i,1>) The second eigenpair from a conjugate pair of eigenvalues is always found from THEOREM. If (lambda1,v1) is an eigenpair of A, then so is (lambda2,v2) where lambda2=conjugate of lambda1 and v2 is the conjugate of v1. EXAMPLE. Find all eigenpairs for A:=Matrix([[2,3],[0,2]]). ANSWER: (2,<1,0>) and there are no others NUMBER of EIGENPAIRS An nxn matrix can have from 1 to n eigenpairs. In the count of eigenpairs, if pairs (lambda1,v1) and (lambda2,v2) are reported, then v1 and v2 must be both nonzero and independent. In any report of eigenpairs, the collection of eigenvectors forms an independent set, although it may be incomplete (less than n eigenvectors). EXAMPLE. Find all eigenpairs for the 3x3 matrix A:=Matrix([[3,0,0],[-4,6,2],[16,-15,-5]]). ANSWER: (0,<0,1,-3>), (1,<0,2,-5>), (3,<1,0,2>) This is an example with 3 distinct eigenvalues. THEOREM. If an eigenvalue lambda is found from the characteristic equation, then there is at least one eigenpair (lambda,v) to report. If all eigenvalues are distinct, then there are n eigenpairs (a full report). EIGENSPACES of a MATRIX A The eigenspace for eigenvalue lambda is the set of all solutions to the equation (A - lambda I)v=0. It is a subspace. ALGORITHM for an EIGENSPACE To find a basis for an eigenspace corrsponding to eigenvalue lambda, proceed as follows. 1. Solve for all eigenpairs (lambda,v). 2. The basis is the set of all v so found, which must be independent by construction. EXAMPLE. Find a basis for each possible eigenspace of A:= Matrix([[4,-2,1],[2,0,1],[2,-2,3]]). ANSWER: eigenspace(lambda=2)=span(<1,0,-2>,<1,1,0>), eigensppace(lambda=3)=span(<1,1,1>). DETAILS: The eigenpairs are (2,<1,0,-2>), (2,<1,1,0>), (3,<1,1,1>). Tedious to find because the characteristic equation is cubic and not easy to factor without computer assist. The book briefly discusses the Theory of Equations from college algebra, which applies to find the roots 2,2,3, IMPORTANT EXERCISES See problems 33, 34, 35, 36, 37, 38, 39, 40, 41, 42. To know eigenanalysis requires knowing all these facts. 6.2 Diagonalization of Matrices ==== EQUATION AP=PD The equation AP=PD is a matrix-multiply reformulation of the n eigenpair relations Av=lambda v, in the special case when the nxn matrix A has exactly eigenpairs. THEOREM. Let the nxn matrix A have a full set of n eigenpairs (lambda,v). Assemble two matrices from the eigenpair report: D = diagonal matrix of eigenvalues = diagonal(lambda_1, ..., lambda_n) P = Matrix of size nxn whose columns are the corresponding eigenvectors = Then matrix multiply implies AP=PD and P is invertble. THEOREM. Assume A is nxn, D is a diagonal matrix, P is nxn invertible and AP=PD. Then the diagonal of D contains the eigenvalues lambda of matrix A and the columns of P contain the corresponding eigenvectors of A. THEOREM. If AP=PD and P is invertible, then A=PDP^(-1). THEOREM. If n eigenpairs of A are known, then D, P are known and then A can be recovered as A=PDP^(-1) EXAMPLE. Let A:=Matrix([[5,-6],[2,-2]]) with eigenpairs (2,<1,-1>), (1,<3,2>). Then AP=PD with D=diag(2,1) and P=<<1,-1>|<3,2>>. SIMILARITY and DIAGONALIZATION. DEF. A, B are SIMILAR if and only if there is an invertible matrix P such that B=P^(-1)AP, or equivalently, AP=PB. DEF. Matrix A is DIAGONALIZABLE if and only if A is nxn and A has exactly n eigenpairs. This means that the eigenvectors reported in the eigenpairs (lambda_k,vk) are linearly independent and the matrix of eigenvectors P= is invertible. EXAMPLE. A:=Matrix([[3,0,0],[-4,6,2],[16,-15,-5]]) has eigenpairs (3,<1,0,2>), (1,<0,2,-5>), (0,<0,1,-3>) and A is diagonalizable. THEOREM 2. If (lambda1,v1) and (lambda2,v2) are eigenpairs with lambda1 different from lambda2, then v1 and v2 are independent. This result extends to independence of k eigenvectors for coreesponding k distinct eigenvalues. THEOREM 3. If nxn matrix A has n distinct eigenvalues, then the corresponding eigenvectors are independent and A is diagonalizable. EXAMPLE. Find all eigenpairs of the 3x3 matrix A:=Matrix([[4,-2,1],[2,0,1],[2,-2,3]]). ANSWER: (2,<1,1,0>), (2,<1,0,-2>), (3,<1,1,1>). The eigenvectors by construction are independent and their augmented matrix P is invertible. THEOREM 4. Let nxn matrix A have n distinct eigenvalues. Then the eigenspaces of A have bases whose union is a complete basis of n eigenvectors of A. ====== end