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## Linear Algebra

by: Dorothea Bode

5

0

9

# Linear Algebra MA 26500

Dorothea Bode
Purdue
GPA 3.97

Staff

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COURSE
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Staff
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9
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KARMA
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## Popular in Mathematics (M)

This 9 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 26500 at Purdue University taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/208119/ma-26500-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15
Supplementary notes for Math 265 on complex eigenvalues eigenvectors and systems of differential equations Clarence Wilkerson In the following we often write the the column vector Z as ab to save space If the 2 gtlt 2 matrix A has distinct real eigenvalues A1 and A2 with corresponding eigenvectors x71 and 72 then the system x t Axt has general solution predicted by the elgenvalue elgenvector method of Agt 7 v1 028 V where the constants 01 and Cg can be determined from the initial values If the matrix A does not have distinct real eigenvalues there can be complications 1 The rst complication is that A need not have any real eigenvalues or eigenvectors This is the topic of these notes 2 A does not have enough eigenvectors we cannot pick a basis for 2 space consisting of 0 1 eigenvectors of A This happens for matrices as simple as 0 0 This complication is discussed brie y in the appendix Main Example of complex conjugate eigenvalues The system docalt y dydt 795 provides the matrix 0 1 A 7 71 0 l The characteristic polynomial is pAgt A2 1 with it In this case the eigenvectors have complex components 7 for A1 239 x71 7t1 For A 7239 one can take x72 t 1 Similar examples occur for A the 2 gtlt 2 matrix expressing the effect of a rotation by angle 6 in the plane namely A 7 cos6 sin6 T 7 sin6 cos6 39 For the rotation above the characteristic equation is pAgt A2 7 2A cos6 1 which has roots cos6 i tsin6 The Main example corresponds to 6 7T2 In general if A is an n gtlt n real matrix pAgt has real coef cients but the eigenvalues and eigenvectors need not be real However the non real eigenvalues and eigenvectors occur in complex conjugate pairs just as in the Main example Theorem Let A be an n gtlt n real matrix Then a if A a tb is an eigenvalue of A then so is the complex conjugate A a 7 tb b if n is a non zero complex vector such that Ax M7 then the complex conjugate of 7 3 1 is an eigenvector for A and the eigenvalue X That is A X Proof Apply complex conjugation to AV and use that A is unchanged by the conjugation since A is real In more detail since Ax M7 one has E 23 AR X lmitating the eigenvalue eigenvector method of the real eigenvalue case leads in the complex case to expressions for the solutions like 05 eaibt 7 Here the exponential is complex and most likely so are the components of 7 If one has started with a real 7 problem like dxdt ydydt 795 this is disconcerting The link back to real solutions is provided by Proposition Euler eltgib eatcosbt isinbt In particular Rae cosa and 7718 sina That is cosa 8 67m27 sina 6 C 7 Tim2239 One can check Euler7s formula by comparing the power series expansions about cc 0 for the exponential sine and cosine functions Going back to the main example dacdt ydydt 795 the eigenvalue eigenvector method predicts a general solution of the form cleit7i 1 Cge it 17 but how does one produce real valued solutions Theorem Suppose A is a square real matrix with complex eigenvalue A and complex eigenvector 7 Then 05 a is a solution to X A Also the complex conjugate a of 3 is a solution In addition the real and imaginary parts of are solutions Proof As in the real case 8 A8 A8 so it is a solution Since A is real so is WV Since it is a linear system the sum of two solutions is again a solution so R80 is a solution and similarly for Warning This depends on A being a matrix with real entries lt7s false without this assumption For our example dacdt ydydt 795 using A 239 yields the complex solution eit7z 1 costisint7i1 sintcost 7 cost7 sint So I Ree t7i 1 1 sintcost7 and I Ime t7i 1 2 7 costsint are also solutions Notice that the Wronskian in this example is det l7 2 sin2t cos2t 1 Note that given complex numbers 2 a M w c dd Re2w ac 7 bd7 Im2w ad bc However7 in practice it may be easier to just multiply out and collect terms than to remember and use this fact If we are given real initial values7 eg i0 1707 then this set of real solutions may be more convenient to work with to solve the initial value problem In this example7 i bl l bg g so fort 0 7 0 170 61071 bg71707 which implies that bl 0 and 6 71 We could also use the complex solutions to solve the same initial value problem 7 put Z1 e 7i71 and Z2 e z 71 Then i clil c2212 and 0 170 017i7102i71 That is7 ficl ic2 17 c1 c2 07 or c1 i027 and c2 12i Hence c1 7122 Exercises 1 1 i i i a a 1 For A 71 1 give two independent real solutions to the system x Ax 2 For B i i analyse as above Also7 nd a solution X such that 0 17 2 3 For the initial value problem at the end of this section7 check that the solution in terms of the complex exponentials agrees with the solution using sines and cosines Graphical interpretation of second order systems The existence of complex eigenvalues may seem arti cal and not related to the physical behavior of the system In fact however7 the Main example is just a linearized version of the second order equation for an harmonic oscillator7 y y 0 We also provide a few graphical examples here to suggest that there are direct consequences of complex eigenvalues From the system AX it is possible with MatLab7 Maple7 or Mathematica to generate pictures that show the behavior of the solution over time We used the Matlab routine pplane from Polking7s book to generate the following pictures In these pictures7 the small arrows give the direction eld 7 at each point 5007740 with a small line with slope dyda and tail at 507 go The solid lines give sample trajectories of solutions that start with 5007 go as initial conditions Unfortunately7 time direction is not indicated on the trajectories lt agrees with the nearby direction elds7 however Case 1 A has distinct real eigenvalues7 one positive and one negative7 eg dacdt fa anddydtyA 01 In this case the eigenvalues are i1 and 00 is a saddle point Notice the con uence of trajectories along x and y axes which are the eigenvectors for this example Sorne direction elds go outward and some go inward Other cases with distinct real eigenvectors are similar If both are negative the trajectories ow in to 00 and 00 is a sink If both eigenvalues are positive then the trajectories ow outward and 07 0 is a source Case 2 A has repeated eigenvalue 72 but not enough eigenvectors dxdt 7295 y and dydt 72y Notice that some trajectories cluster along the eigen vectorthe x axis and that the direction elds point inward because A has negative real part Case 3 Purely irnaginary eigenvalues ie dxdt 2y and dydt 72a Notice the elliptical trajectories This case corresponds to the harmonic oscillator equation with no friction Case 4 Complex with negative real values dxdt 7295 2g and dydt 72a 7 2y Notice that the trajectories and direction elds swirl inward 77 If the real part of the eigenvalues were positive7 then the trajectories and direction elds would swirl outward 7 Appendix A glimpse of the repeated eigenvalue problem If the n gtlt 71 matrix is such that one can nd n Iinearly independent vectors x77 which are eigenvectors for A then we say that A has enough eigenvectors or that A is diagonalizable In this case the eigenvalue eigenvecor method produces a correct general solution to i A Namely i 231 cjekjtvj 2 0 For example if A 0 2 then 10 and 01 are eigenvectors for A 2 and the solution is a linear combination of 82t1 0 and 827 1 However ifA has repeated eigenvalues there need not be a basis for 71 or C consisting of eigenvectors of A In this case the naive77 eigenvalue eigenvector method fails Suppose that A with A real Then A has repeated eigenvalues A and A A 0 However multiples of V1 10 are the only eigenvectors In this case the eigenvalue eigenvector method gives only solutions to X A of the form cemvl However the general theory predicts that for A a 2 gtlt 2 matrix one should have two linearly independent solutions Again linear algebra provides the missing solution Proposition For A as above if there are non zero vectors x71 and V such that AiA71 0 and A 7 A72 v1 then a x71 and V are linearly independent and b envl and 8M 15 72 are linearly independent solutions to 3 A Proofa If AiA72 x71 and AiA71 0 then if 61 11 0 apply AiA to get a0b71 0 If x71 3 0 then b 0 and hence a 0 That is 71 and V are linearly independent Proofb 8 15x71 my er AteAtvl Aektvg A AeM 15x71 72 eMtA71 eAtAWg 5er er MW Comparing terms shows that i Ax Also putting t 0 in er yields 71 and similarly X7 in the second solution Since 121 and 122 are independent by a the solutions are linearly independent 0 A In this example this leads to row reducing 0 1 1 which has among its solutions A 1 01 Thus forAiAi 0 A 8 151 0 01 ten 8 Notice that the Wronskian in this case is 82 3 0 For the example A A 1 x71 10 Then one must solve A 7 A72 x71 two independent solutions are e 10 8M70 and More complicated examples ofmatrices with repeated eigenvalues are discussed in the section on Jordan forms in Math 511 The solutions to the associated differential equation is analo gous to the solutions to higher order linear constant coef cient ODE s with repeated roots There is in principle a method that avoids the problems of the eigenvalue eigenvector method Put 8 Id 2130114071711 This series converges for all matrices A Then 05 eAtVO is a solution to the system t A t with initial value 70 In practice7 computing 8 is often unwieldy and best performed using eigenvalue eigenvector methods However7 it does a provide some insight to the solution in the repeated root case For A 0 7 one can eat teat 0 eat This topic also is covered in Math 511 easily show that 8 Exercises 4 Ifgt 07 so that A 8 1 verify that Ati 1t 8 01 HintA20 010 5lfgt0andB 0 0 1 7verifythat 0 0 0 1tt22 837 012 0 01 Hint 33 07 32 7 6 Suppose that there are non zero vectors x77 for 0 lt j S k such that A 7 A7j X774 interpreting x70 0 Show that k 05 8At7k Jr N164 Jr tgQl7k2 Jr ts3l7k3 Jr 8M ZtkijU 7 j1 is a solution to Homogeneous systems Remember A homogeneous system always has at least one solution the trivial solution x 0 If it has another solution then it has in nitely many of them a Whole subspace of them because we can multyply one of the solution by any constant and we get a different solution Homework Section 15

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