DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

More documents

Recommendations

Info

100 II. DETERMINANTS AND EIGENVALUESis of the form a + bλ. Show by using the recursive definition of the determinantthat det B(λ) is a polynomial in λ of degree at most n. Now use this fact and therecursive definition of the determinant to show that det(A − λI) is a polynomial ofdegree exactly n.9. (Project) The purpose of this project is to illustrate one method for approximatingan eigenvector and the corresponding eigenvalue in cases where exact calculationis not feasible. We use an example in which one can find exact answers bythe usual method, at least if one uses radicals, so we can compare answers to gaugehow effective[ the method is.1 1Let A =1 2Let v 0 =]. Define an infinite sequence of vectors v n in R 2 as follows.[10]. Having defined v n , define v n+1 = Av n . Thus, v 1 = Av 0 , v 2 =Av 1 = A 2 v 0 , v 3 = Av 2 = A 3 v 0 , etc. Then it turns out in this case that asn →∞, the directions of the vectors v n approach the direction of an eigenvectorfor A. Unfortunately, there is one difficulty: the magnitudes |v n | approach [ ] infinity.anTo get get around this problem, proceed as follows. Let v n = and putb nu n = 1 [ ]an /bv n = n. Then the second component is always one, and the firstb n 1[ ]rcomponent r n = a n /b n approaches a limit r and u = is an eigenvector for A.1(a) For the above matrix A, calculate the sequence of vectors v n and and numbersr n n =1,2,3,.... Do the calculations for enough n to see a pattern emerging andso that you can estimate r accurately to 3 decimal places.(b) Once you know an eigenvector u, you can find the corresponding eigenvalueλ by calculating Au. Use your estimate in part (a) to estimate the correspondingλ.(c) Compare this to the roots of the characteristic equation (1−λ)(2−λ)−1 =0.Note that the method employed here only gives you one of the two eigenvalues. Infact, this method, when it works, usually gives the largest eigenvalue.5. DiagonalizationIn many cases, the process outlined in the previous section results in a basis forR n which consists of eigenvectors for the matrix A. Indeed, the set of eigenvectorsso obtained is always linearly independent, so if it is large enough (i.e., has nelements), it will be a basis. When that is the case, the use of that basis toestablish a coordinate system for R n can simplify calculations involving A.Example 1. Let[ ]2 1A = .1 2We found in the previous section that[ ] [ ]1 −1v 1 = , v1 2 =1
5. DIAGONALIZATION 101are eigenvectors respectively with eigenvalues λ 1 = 3 and λ 2 = 1. The set {v 1 , v 2 }is linearly independent, and since it has two elements, it must be a basis for R 2 .Suppose v is any vector in R 2 . We may express it respect to this new basis[ ]y1v = v 1 y 1 + v 2 y 2 =[v 1 v 2 ]y 2where (y 1 ,y 2 ) are the coordinates of v with respect to this new basis. It followsthatAv = A(v 1 y 1 + v 2 y 2 )=(Av 1 )y 1 +(Av 2 )y 2 .However, since they are eigenvectors, each is just multiplied by the correspondingeigenvalue, or in symbolsAv 1 = v 1 (3) and Av 2 = v 2 (1).So(1) A(v 1 y 1 + v 2 y 2 )=v 1 (3y 1 )+v 2 y 2 =[v 1[ ]3y1v 2 ] .y 2In other words, with respect to the new coordinates, the effect of multiplication byA on a vector is to multiply the first new coordinate by the first eigenvalue λ 1 =3and the second new coordinate by the second eigenvalue λ 2 =1.Whenever there is a basis for R n consisting of eigenvectors for A, we say that Ais diagonalizable and that the new basis diagonalizes A.The reason for this terminology may be explained as follows. In the aboveexample, rewrite the left most side of equation (1)[ ]y1A(v 1 y 1 + v 2 y 2 )=A[v 1 v 2 ]y 2and the right most side as[ ]3y1[ v 1 v 2 ] =[vy 1 v 2 ]2[ ][3 0 y10 1y 2].]Since these two are equal, if we drop the common factor on the right, we gety 2[ ]3 0A [ v 1 v 2 ]=[v 1 v 2 ] .0 1Let[ ]1 −1P =[v 1 v 2 ]= ,1 1i.e., P is the 2 × 2 matrix with the basic eigenvectors v 1 , v 2 as columns. Then, theabove equation can be writtenAP = P[ ]3 00 1or P −1 AP =[y1[ ]3 0.0 1
Page 4 and 5: 74 II. DETERMINANTS AND EIGENVALUES
Page 6 and 7: 76 II. DETERMINANTS AND EIGENVALUES
Page 8: 78 II. DETERMINANTS AND EIGENVALUES
Page 11 and 12: 2. DEFINITION OF THE DETERMINANT 81
Page 13 and 14: 3. SOME IMPORTANT PROPERTIES OF DET
Page 21 and 22: 4. EIGENVALUES AND EIGENVECTORS FOR
Page 29: 4. EIGENVALUES AND EIGENVECTORS FOR
Page 33 and 34: 5. DIAGONALIZATION 103Example 3. Co
Page 35 and 36: 6. THE EXPONENTIAL OF A MATRIX 1053
Page 37 and 38: 6. THE EXPONENTIAL OF A MATRIX 107(
Page 39 and 40: 7. REVIEW 109⎡3. A = ⎣ 0 1 1⎤

DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

Create successful ePaper yourself

Delete template?

Save as template?