Chapter EIGENVALUES AND EIGENVECTORS PDF document - DocSlides

Chapter  EIGENVALUES AND EIGENVECTORS PDF document - DocSlides

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1 Motivation We motivate the chapter on eigenvalues by discussing the equ ation ax 2 hxy by c where not all of a h b are zero The expression ax 2 hxy by is called quadratic form in and and we have the identity ax 2 hxy by x y a h h b AX where an ID: 24033

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Presentations text content in Chapter EIGENVALUES AND EIGENVECTORS


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Chapter 6 EIGENVALUES AND EIGENVECTORS 6.1 Motivation We motivate the chapter on eigenvalues by discussing the equ ation ax + 2 hxy by c, where not all of a, h, b are zero. The expression ax + 2 hxy by is called quadratic form in and and we have the identity ax + 2 hxy by x y a h h b AX, where and a h h b is called the matrix of the quadratic form. We now rotate the x, y axes anticlockwise through radians to new , y axes. The equations describing the rotation of axes are deri ved as follows: Let have coordinates ( x, y ) relative to the x, y axes and coordinates , y ) relative to the , y axes. Then referring to Figure 6.1: 115
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116 CHAPTER 6. EIGENVALUES AND EIGENVECTORS @ Figure 6.1: Rotating the axes. OQ OP cos ( OP (cos cos sin sin = ( OP cos )cos OP sin )sin OR cos PR sin cos sin θ. Similarly sin cos We can combine these transformation equations into the sing le matrix equation: cos sin sin cos or PY , where , Y and cos sin sin cos We note that the columns of give the directions of the positive and axes. Also is an orthogonal matrix – we have PP and so The matrix has the special property that det = 1. A matrix of the type cos sin sin cos is called a rotation matrix. We shall show soon that any 2 2 real orthogonal matrix with determinant
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6.1. MOTIVATION 117 equal to 1 is a rotation matrix. We can also solve for the new coordinates in terms of the old on es: cos sin sin cos so cos sin and sin cos . Then AX = ( PY PY ) = AP Y. Now suppose, as we later show, that it is possible to choose an angle so that AP is a diagonal matrix, say diag( , ). Then AX (6.1) and relative to the new axes, the equation ax + 2 hxy by becomes , which is quite easy to sketch. This curve is symmetrical about the and axes, with and , the respective columns of giving the directions of the axes of symmetry. Also it can be verified that and satisfy the equations AP and AP These equations force a restriction on and . For if , the first equation becomes a h h b or h b Hence we are dealing with a homogeneous system of two linear e quations in two unknowns, having a non–trivial solution ( , v ). Hence h b = 0 Similarly, satisfies the same equation. In expanded form, and satisfy ab = 0 This equation has real roots 4( ab + 4 (6.2) (The roots are distinct if or = 0. The case and = 0 needs no investigation, as it gives an equation of a circle.) The equation ab = 0 is called the eigenvalue equation of the matrix
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118 CHAPTER 6. EIGENVALUES AND EIGENVECTORS 6.2 Definitions and examples DEFINITION 6.2.1 (Eigenvalue, eigenvector) Let be a complex square matrix. Then if is a complex number and non–zero com- plex column vector satisfying AX λX , we call an eigenvector of while is called an eigenvalue of . We also say that is an eigenvector corresponding to the eigenvalue So in the above example and are eigenvectors corresponding to and , respectively. We shall give an algorithm which starts from t he eigenvalues of a h h b and constructs a rotation matrix such that AP is diagonal. As noted above, if is an eigenvalue of an matrix , with corresponding eigenvector , then ( λI = 0, with = 0, so det( λI ) = 0 and there are at most distinct eigenvalues of Conversely if det ( λI ) = 0, then ( λI = 0 has a non–trivial solution and so is an eigenvalue of with a corresponding eigenvector. DEFINITION 6.2.2 (Characteristic polynomial, equation) The polynomial det ( λI ) is called the characteristic polynomial of and is often denoted by ch ). The equation det( λI ) = 0 is called the characteristic equation of . Hence the eigenvalues of are the roots of the characteristic polynomial of For a 2 2 matrix a b c d , it is easily verified that the character- istic polynomial is (trace +det , where trace is the sum of the diagonal elements of EXAMPLE 6.2.1 Find the eigenvalues of 2 1 1 2 and find all eigen- vectors. Solution . The characteristic equation of is + 3 = 0, or 1)( 3) = 0 Hence = 1 or 3. The eigenvector equation ( λI = 0 reduces to 1 2
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6.2. DEFINITIONS AND EXAMPLES 119 or (2 = 0 + (2 = 0 Taking = 1 gives = 0 = 0 which has solution y, y arbitrary. Consequently the eigenvectors corresponding to = 1 are the vectors , with = 0. Taking = 3 gives = 0 = 0 which has solution y, y arbitrary. Consequently the eigenvectors corre- sponding to = 3 are the vectors , with = 0. Our next result has wide applicability: THEOREM 6.2.1 Let be a 2 2 matrix having distinct eigenvalues and and corresponding eigenvectors and . Let be the matrix whose columns are and , respectively. Then is non–singular and AP Proof . Suppose AX and AX . We show that the system of homogeneous equations xX yX = 0 has only the trivial solution. Then by theorem 2.5.10 the mat rix ] is non–singular. So assume xX yX = 0 (6.3) Then xX yX ) = 0 = 0, so AX ) + AX ) = 0. Hence x y = 0 (6.4)
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120 CHAPTER 6. EIGENVALUES AND EIGENVECTORS Multiplying equation 6.3 by and subtracting from equation 6.4 gives yX = 0 Hence = 0, as ( = 0 and = 0. Then from equation 6.3, xX = 0 and hence = 0. Then the equations AX and AX give AP ] = [ AX AX ] = [ = [ so AP EXAMPLE 6.2.2 Let 2 1 1 2 be the matrix of example 6.2.1. Then and are eigenvectors corresponding to eigenvalues 1 and 3, respectively. Hence if 1 1 1 1 , we have AP 1 0 0 3 There are two immediate applications of theorem 6.2.1. The fir st is to the calculation of : If AP = diag ( , ), then diag ( , and The second application is to solving a system of linear differe ntial equations dx dt ax by dy dt cx dy, where a b c d is a matrix of real or complex numbers and and are functions of . The system can be written in matrix form as AX where and dx dt dy dt
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6.2. DEFINITIONS AND EXAMPLES 121 We make the substitution PY , where . Then and are also functions of and AX PY so = ( AP Y. Hence and These differential equations are well–known to have the solut ions (0) and (0) , where (0) is the value of when = 0. [If dx dt kx , where is a constant, then dt kt ke kt kt dx dt ke kt kt kx = 0 Hence kt is constant, so kt (0) = (0). Hence (0) kt .] However (0) (0) (0) (0) , so this determines (0) and (0) in terms of (0) and (0). Hence ultimately and are determined as explicit functions of , using the equation PY EXAMPLE 6.2.3 Let . Use the eigenvalue method to derive an explicit formula for and also solve the system of differential equations dx dt = 2 dy dt = 4 y, given = 7 and = 13 when = 0. Solution . The characteristic polynomial of is +3 +2 which has distinct roots 1 and 2. We find corresponding eigenvectors and . Hence if 1 3 1 4 , we have AP = diag ( 2). Hence diag ( 2) diag (( 1) 2) 1 3 1 4 1) 0 ( 2) 1 1
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122 CHAPTER 6. EIGENVALUES AND EIGENVECTORS = ( 1) 1 3 1 4 1 0 0 2 1 1 = ( 1) 1 3 1 4 1 1 = ( 1) 3 + 3 3 + 4 To solve the differential equation system, make the substitut ion PY . Then + 3 , y + 4 . The system then becomes so (0) , y (0) . Now (0) (0) (0) (0) 1 1 13 11 so 11 and = 6 . Hence 11 + 3(6 ) = 11 18 , y 11 + 4(6 ) = 11 + 24 For a more complicated example we solve a system of inhomogeneous recurrence relations. EXAMPLE 6.2.4 Solve the system of recurrence relations +1 = 2 +1 + 2 + 2 given that = 0 and 1. Solution . The system can be written in matrix form as +1 AX B, where 1 2 and It is then an easy induction to prove that + ( B. (6.5)
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6.2. DEFINITIONS AND EXAMPLES 123 Also it is easy to verify by the eigenvalue method that 1 + 3 1 + 3 V, where 1 1 1 1 and 1 1 . Hence (3 + 3 + 1) (3 1) V. Then equation 6.5 gives (3 1) which simplifies to (2 + 1 (2 5 + 3 Hence = (2 + 1 4 and = (2 5 + 3 4. REMARK 6.2.1 If ( existed (that is, if det( = 0, or equivalently, if 1 is not an eigenvalue of ), then we could have used the formula = ( )( (6.6) However the eigenvalues of are 1 and 3 in the above problem, so formula 6.6 cannot be used there. Our discussion of eigenvalues and eigenvectors has been lim ited to 2 matrices. The discussion is more complicated for matrices o f size greater than two and is best left to a second course in linear algebra. Nevertheless the following result is a useful generalization of theorem 6 .2.1. The reader is referred to [28, page 350] for a proof. THEOREM 6.2.2 Let be an matrix having distinct eigenvalues , . . ., and corresponding eigenvectors , . . ., X . Let be the matrix whose columns are respectively , . . ., X . Then is non–singular and AP 0 0
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124 CHAPTER 6. EIGENVALUES AND EIGENVECTORS Another useful result which covers the case where there are m ultiple eigen- values is the following (The reader is referred to [28, pages 351–352] for a proof): THEOREM 6.2.3 Suppose the characteristic polynomial of has the fac- torization det( λI ) = ( where , . . ., c are the distinct eigenvalues of . Suppose that for , . . ., t , we have nullity ( ) = . For each such , choose a basis , . . ., X in for the eigenspace ). Then the matrix = [ 11 || || || tn is non–singular and AP is the following diagonal matrix AP 0 0 (The notation means that on the diagonal there are elements , followed by elements ,. .. , elements .) 6.3 PROBLEMS 1. Let 1 0 . Find an invertible matrix such that AP diag (1 3) and hence prove that 2. If 6 0 4 0 , prove that tends to a limiting matrix 3 2 3 1 as
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6.3. PROBLEMS 125 3. Solve the system of differential equations dx dt = 3 dy dt = 5 y, given = 13 and = 22 when = 0. [Answer: = 7 + 6 , y = 7 + 15 .] 4. Solve the system of recurrence relations +1 = 3 +1 + 3 given that = 1 and = 2. [Answer: = 2 (3 , y = 2 (3 + 2 ).] 5. Let a b c d be a real or complex matrix with distinct eigenvalues , and corresponding eigenvectors , X . Also let = [ ]. (a) Prove that the system of recurrence relations +1 ax by +1 cx dy has the solution where and are determined by the equation (b) Prove that the system of differential equations dx dt ax by dy dt cx dy has the solution αe βe
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126 CHAPTER 6. EIGENVALUES AND EIGENVECTORS where and are determined by the equation (0) (0) 6. Let 11 12 21 22 be a real matrix with non–real eigenvalues ib and ib , with corresponding eigenvectors iV and iV , where and are real vectors. Also let be the real matrix defined by = [ ]. Finally let ib re i , where r > 0 and is real. (a) Prove that AU aU bV AV bU aV. (b) Deduce that AP a b b a (c) Prove that the system of recurrence relations +1 11 12 +1 21 22 has the solution αU βV )cos n + ( βU αV )sin n where and are determined by the equation (d) Prove that the system of differential equations dx dt ax by dy dt cx dy
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6.3. PROBLEMS 127 has the solution at αU βV )cos bt + ( βU αV )sin bt where and are determined by the equation (0) (0) [Hint: Let . Also let iy . Prove that = ( ib and deduce that iy at i )(cos bt sin bt Then equate real and imaginary parts to solve for , y and hence x, y .] 7. (The case of repeated eigenvalues.) Let a b c d and suppose that the characteristic polynomial of + ( ad bc ), has a repeated root . Also assume that αI . Let αI (i) Prove that ( + 4 bc = 0. (ii) Prove that = 0. (iii) Prove that BX = 0 for some vector ; indeed, show that can be taken to be or (iv) Let BX . Prove that = [ ] is non–singular, AX αX and AX αX and deduce that AP 8. Use the previous result to solve system of the differential e quations dx dt = 4 dy dt = 4 + 8 y,
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128 CHAPTER 6. EIGENVALUES AND EIGENVECTORS given that = 1 = when = 0. [To solve the differential equation dx dt kx , k a constant multiply throughout by kt , thereby converting the left–hand side to dx dt kt ).] [Answer: = (1 , y = (1 + 6 .] 9. Let 2 1 2 0 4 1 4 1 4 1 4 1 (a) Verify that det( λI ), the characteristic polynomial of , is given by 1) (b) Find a non–singular matrix such that AP = diag (1 ). (c) Prove that 1 1 1 1 1 1 1 1 1 2 2 1 2 1 2 if 1. 10. Let 5 2 2 5 2 5 (a) Verify that det( λI ), the characteristic polynomial of , is given by 3) 9) (b) Find a non–singular matrix such that AP = diag (3 9).

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