Linear Algebra: Matrix Eigenvalue Problems

Transcription

1 Chap 8 Linear Algebra: Matrix Eigenvalue Problems Sec 81 Eigenvalues, Eigenvectors Eigenvectors of a (square!) matrix A are vectors x, not zero vectors, such that if you multiply them by A, you get a vector y Ax that is proportional to x, say, y x The factor of proportionalityis called an eigenvalue of A How to find the eigenvalues? You have to solve the characteristic equation (4) of A, which is a quadratic equation in if A is a 22matrix, a cubic equation for a matrix, and so on Hence in practice you may need a root-finding method or one of the interesting iterative methods to be discussed in numerics in Chap 20 Once you have found an eigenvalue, to find a corresponding eigenvector is simpler You can find it by solving a linear system of equations, as explained on p 7 and later in this chapter Or you get an eigenvector along with an eigenvalue by using those iterative methods just mentioned Problem Set 81 Page 8 1 Eigenvalues and eigenvectors For a diagonal matrix the eigenvalues are the main diagonal entries because the characteristic equation is detai a a 22 a 11 a 22 0 For the given matrix you obtain from this 1 2, 2 04 Now determine an eigenvector of A corresponding to 1 2 In components, A 1 Ix 0 is a 11 1 x 1 a 12 x 2 2 2x 1 0x 2 0 a 21 x 1 a 22 1 x 2 0x 1 042x 2 0 The first equation gives no condition The second gives x 2 0 Hence an eigenvector of A corresponding to 1 2 isx 1 0 T Since an eigenvector is determined only up to a nonzero constant, you can simply take1 0 T as an eigenvector For 2 04 the procedure is similar and leads to 0 1 T 5 Eigenvalues and eigenvectors Problem 1 concerned a diagonal matrix, a case in which you could see the eigenvalues almost immediately For a general 2 2 matrix the determination of eigenvalues and eigenvectors follows the same pattern Example 1 on p 4 illustrates this For Prob 5 the matrix is Calculate the characteristic equation A and from it the eigenvalues 1 4, 2 Then find eigenvectors For 1 4 obtain the system (2), p 5,

2 80 Linear Algebra, Vector Calculus Part B 5 4x 1 2x 2 0 say, x 1 2, x 2 9 9x 1 6 4x 2 0 (not needed) You thus have the eigenvector x T corresponding to 1 4 For 2 obtain the system (2) 5x 1 2x 2 0 say, x 1 1, x 2 1 9x 1 6 x 2 0 (not needed) You thus have the eigenvector x T corresponding to 2 Keep in mind that eigenvectors are determined only up to a nonzero constant factor 19 Eigenvalues and eigenvectors Ordinarily one would expect that a matrix has linearly independent eigenvectors For symmetric, skew-symmetric and many other matrices this is true A simple example is the unit matrix, which has but one eigenvalue, 1, but every (nonzero) vector is an eigenvector, so that you can choose, for instance, T,0 1 0 T, T The given matrix has the characteristic equation Conclude that 9 is an eigenvalue of algebraic multiplicity Find eigenvectors Set9in the characteristic matrix to get Row-reduce it by Gauss to get A9I You see that it has rank 2, so you can choose one unknown (one component of an eigenvector) and then determine the other two components Say, x 1 Then find x 2 2 from the second row, 9 2 x 2 9x 0, and finally x x 2 2x 2 from the first row Hence you found T, and the defect is 2 (the geometric multiplicity is 1 because you cannot find another linearly independent eigenvector) 29 Complex eigenvectors In more detail, the answer is as follows Since the matrix is real, which by definition means that its entries are all real, the coefficients of the characteristic polynomial are real, and you know from algebra that a polynomial with real coefficients has real roots or roots that are complex conjugate in pairs

3 Chap 8 Linear Algebra: Matrix Eigenvalue Problems 81 Sec 82 Some Applications of Eigenvalue Problems Problem Set 82 Page 4 17 Open Leontief input-output model For reasons explained in the statement of the problem you have to solve x Ax y for x, where A and y are given With the given data you thus have to solve x Ax IAx x 1 x 2 x y For this you can apply the Gauss elimination to the augmented matrix of the system If you use 6 decimals in your calculation, you will obtain the solution (rounded) x 1 07, x 2 059, x Markov process In Example 2 on p 41 you considered the transpose because from it you could immediately see that it has the eigenvalue 1 and T as a corresponding eigenvector In the present problem the situation is even simpler because the given matrix is symmetric and you can immediately conclude that it has the eigenvalue 1 with T as a corresponding eigenvector, which characterizes the limit state Sec 8 Symmetric, Skew-Symmetric, and Orthogonal Matrices Example 1 Notice that the defining properties (1) and (2) can be seen immediately In particular, for a skew-symmetric matrix you have a jj a jj, hence the main diagonal entries must be 0 Problem Set 8 Page 48 5 Inverse of a skew-symmetric matrix Let A be skew-symmetric, that is, A T A Let A be nonsingular Let B be its inverse Then AB I Transposition of (2) and the use of the skew symmetry (1) of A give I I T AB T B T A T B T A B T A Now multiply () by B from the right and use (2), obtaining B B T AB B T This proves that B A 1 is skew-symmetric (1) (2) () 11 A common mistake The matrix 1 A 1 1 is not skew-symmetric because its main diagonal entries are not 0 Show that its characteristic equation is ,

4 82 Linear Algebra, Vector Calculus Part B so that it has the double root 2, whose defect is 1; an eigenvector is1 1 T Similarly, note that the matrix A 1 1 is not skew-symmetric Its characteristic equation is ii 0 The eigenvalues are i Eigenvectors are1 i T and 1 i T Sec 84 Eigenbases Diagonalization Quadratic Forms Content For diagonalizing a matrix you need an eigenbasis (a basis of eigenvectors) Theorems 1 and 2 give the practically most important cases of the existence of a basis Diagonalization is done by a similarity transformation (definition on p 50) with a suitable matrix X constructed from eigenvectors as shown in (5) in Theorem 4 Diagonalization is applied to quadratic forms ( transformation to principal axes ) in Theorem 5 Problem Set 84 Page 55 1 Diagonalization You need the eigenvalues and then eigenvectors of A for making up the matrices X and X 1 by means of which you diagonalize Find the eigenvalues from the characteristic equation Thus 1 7 and 2 2 Now obtain eigenvectors for 1 7 from7x 1 2x 2 4x 1 2x 2 0, say, x 1 1, x 2 2 for 2 2 from2x 1 2x 2 x 1 2x 2 0, say, x 1 2, x 2 1 With these two eigenvectors1 2 T and 2 1 T make up the matrix X and calculate its inverse, X , det X 5, X You are now ready to calculate the product that will give you the diagonal matrix (the result of the diagonalization of A), showing the eigenvalues in the main diagonal, X 1 AX The two eigenvectors that you have determined form an eigenbasis of A 15 Preservation of spectrum Prove that A and  P 1 AP, P 1 6 have the same spectrum You need det P and (use (4*), p 18) P You then obtain

5 Chap 8 Linear Algebra: Matrix Eigenvalue Problems 8  P 1 AP P Show the equality of the eigenvalues by calculating and detâi detai Find an eigenvector y 1 of  for 1 2 from Calculate x 1 72x 2 0, 2x 1 72x 2, say, x 1 9, x 2 4 x 1 Py Calculate an eigenvector y 2 of  for 2 0 from Calculate 0x 1 72x 2 0, say, x 1 12, x 2 5 x 2 Py Sec 85 Complex Matrices and Forms Optional Example 2 In A the diagonal entries are real, hence equal to their conjugates a 21 1i is the complex conjugate of a 12 1i, as it should be for a Hermitian matrix In B you have b 11 i i b 11, b 12 2 i 2i b 21, and b 22 i b 22 The complex conjugate transpose of C is C T i/2 /2 /2 i/2 Multiply this by C to obtain the unit matrix This verifies the defining relationship of a unitary matrix Problem Set 85 Page 61 5 Hermitian matrix a 11 and a 22 are real a 21 i ā 12 Conclude that A is Hermitian Calculate the characteristic equation 42 ii The eigenvalues are 2 Find an eigenvector for 1 2 from say, 4 1 x 1 ix 2 1 2x 1 ix 2 0, x 1 i, x 2 1 i 1 2 i 1 2

6 84 Linear Algebra, Vector Calculus Part B Similarly, find an eigenvector for 2 2 from 4 2 x 1 ix 2 1 2x 1 ix 2 0, say, x 1 i, x 2 1 i 1 2 i Skew-Hermitian form Note that a 11 a 22 0, and a 21 i i a 12 and conclude that A is skew-hermitian Expect the form to have a pure imaginary value or the value 0 Calculate x T Ax 4i i 0 i i 0 4 i i 4i i i i i4 i 4 iii ii4 i 66i