55.1© 2012 Pearson Education, Inc.Eigenvalues and EigenvectorsEIGENVECTORS AND EIGENVALUESSlide 5.1- 2© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Definition: An eigenvector of an matrix A is a nonzero vector x such that for some scalar λ. A scalar λ is called an eigenvalue of A if there is a nontrivial solution x of ; such an x is called an eigenvector corresponding to λ. λ is an eigenvalue of an matrix A if and only if the equation----(1)has a nontrivial solution. The set of all solutions of (1) is just the null space of the matrix .nnx λxAx λxA( λ )x 0AInnλAISlide 5.1- 3© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES So this set is a subspace of and is called the eigenspace of A corresponding to λ. The eigenspace consists of the zero vector and all the eigenvectors corresponding to λ. Example 1: Show that 7 is an eigenvalue of matrixand find the corresponding eigenvectors.n1652ASlide 5.1- 4© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Solution: The scalar 7 is an eigenvalue of A if and only if the equation----(2)has a nontrivial solution. But (2) is equivalent to , or ----(3) To solve this homogeneous equation, form the matrixx7xAx7x0A(7)x0AI16 70 6 6752 07 5 5AI        Slide 5.1- 5© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES The columns of are obviously linearly dependent, so (3) has nontrivial solutions. To find the corresponding eigenvectors, use row operations: The general solution has the form . Each vector of this form with is an eigenvector corresponding to .7AI660 110550 000211x20x λ 7Slide 5.1- 6© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Example 2: Let . An eigenvalue of A is 2. Find a basis for the corresponding eigenspace. Solution: Formand row reduce the augmented matrix for . 416216218A416 200 2162 2 16 020 2 16218 002 216AI   (2)x0AISlide 5.1- 7© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES At this point, it is clear that 2 is indeed an eigenvalue of A because the equation has free variables. The general solution is, x2and x3free.2 160 2 1602 160 0 0002 160 0 000(2)x0AI122 331/ 2 31001xxx xx                    Slide 5.1- 8© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES The eigenspace, shown in the following figure, is a two-dimensional subspace of . A basis is 3132, 001     Slide 5.1- 9© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Theorem 1: The eigenvalues of a triangular matrix are the entries on its main diagonal. Proof: For simplicity, consider the case. If A is upper triangular, the has the form33λAI11 12 1322 233311 12 1322 2333λ 00λ 00λ 000 00λλ0 λ00 λaaaAI a aaaaaaaa      Slide 5.1- 10© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES The scalar λ is an eigenvalue of A if and only if the equation has a nontrivial solution, that is, if and only if the equation has a free variable. Because of the zero entries in , it is easy to see that has a free variable if and only if at least one of the entries on the diagonal of is zero.  This happens if and only if λ equals one of the entries a11, a22, a33in A. ( λ )x 0AIλAI( λ )x 0AIλAISlide 5.1- 11© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Theorem 2: If v1, …, vrare eigenvectors that correspond to distinct eigenvalues λ1, …, λrof anmatrix A, then the set {v1, …, vr} is linearly independent.  Proof: Suppose {v1, …, vr} is linearly dependent. Since v1is nonzero, Theorem 7 in Section 1.7 says that one of the vectors in the set is a linear combination of the preceding vectors. Let p be the least index such that is a linear combination of the preceding (linearly independent) vectors.nn1vpSlide 5.1- 12© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Then there exist scalars c1, …, cpsuch that----(4) Multiplying both sides of (4) by A and using the fact that ----(5) Multiplying both sides of (4) by and subtracting the result from (5), we have----(6)11 1vvvpp pcc11 1111 1 1vvvλ v λ v λ vpp pppp p pcA c A Acc 1λp11 11 1(λλ)v (λλ)v 0pppppcc Slide 5.1- 13© 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUES Since {v1, …, vp} is linearly independent, the weights in (6) are all zero. But none of the factors are zero, because the eigenvalues are distinct. Hence for . But then (4) says that , which is impossible.1λλip0ic  1, ,ip1v0pSlide 5.1- 14© 2012 Pearson Education, Inc.EIGENVECTORS AND DIFFERENCE EQUATIONS Hence {v1, …, vr} cannot be linearly dependent and therefore must be linearly independent.  If A is an matrix, then ----(7) is a recursive description of a sequence {xk} in . A solution of (7) is an explicit description of {xk} whose formula for each xkdoes not depend directly on A or on the preceding terms in the sequence other than the initial term x0.nn1xxkkA(0,1,2)knSlide 5.1- 15© 2012 Pearson Education, Inc.EIGENVECTORS AND DIFFERENCE EQUATIONS The simplest way to build a solution of (7) is to take an eigenvector x0and its corresponding eigenvalue λand let----(8) This sequence is a solution because0x λ