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18.03 Class 34, April 30, 2010 Complex or repeated eigenvalues 1. Eigenvalues and coefficients 2. Complex eigenvalues 3. Repeated eigenvalues 4. Defective and complete [1] We were solving u' = Au , with A = [ a b ; c d ] : A = [ -2 1 ; -4 3 ] for example. Or u could be a 3D vector, and A a 3x3 matrix. (1) Find the eigenvalues = roots lambda_1, lambda_2 of the characteristic polynomial p_A(lambda) = det( A - lambda I ) (2) For each eigenvalue, find a nonzero eigenvector: ( A - lambda_1 I ) v_1 = 0 ( A - lambda_2 I ) v_2 = 0 Normal modes: u_1 = e^{\lambda_1 t} v_1 , u_2 = e^{\lambda_2 t} v_2 . (3) General solution is linear combination of these: u = c_1 u_1 + c_2 u_2 . There are a few problems with this, but before pointing them out let me make three comments: (a) Any multiple of an eigenvector is another eigenvector for the same eigenvalue; they form a line, an "eigenline." (b) The zero vector is an eigenvector for every value lambda, whether lambda is an eigenvalue or not. Most of the time 0 is the ONLY eigenvector for value lambda ; lambda is an eigenvalue exactly when there is a *nonzero* eigenvector for that value. (c) p_A(lambda) = det( A - lambda I ) = lambda^2 - (a+d) lambda - (ad-bc) The sum of the diagonal terms of a square matrix is the "trace" of A , tr A , so p_A(lambda) = lambda^2 - (tr A) lambda + (det A) In our example, tr A = 1 and det A = -2 , and p_A(lambda) = lambda^2 - lambda - 2 . [2] There may be no ray solutions. Romeo and Juliet provided one example. Or, what about A = [ 1 2 ; -2 1 ] .Let's apply the method and see what happens. tr(A) = 2 , det(A) = 5, so p_A(lambda) = lambda^2 - 2 lambda + 5 = (lambda - 1)^2 + 4 which has roots lambda_1 = 1 + 2i, lambda_2 = 1 - 2i. We could abandon the effort at this point, but we had so much fun and success with complex numbers earlier that it seems we should carry on. Find an eigenvector for lambda_1 = 1 + 2i : A - (1+2i)I : [ - 2i 2 ; -2 -2i ][ ? ; ? ] = [ 0 ; 0 ] Standard method: use the entries in the top row in reverse order with one sign changed: [ 2 ; 2i ] or, easier, in this case, v_1 = [ 1 ; i ]. This is set up so the top entry in the product is 0 . We have a chance to check our work (mainly the calculation of the eigenvalues) by seeing that the bottom entry in the product is 0 too: -2 . 1 - 2i . i = 0 [ 1 ; i ] is a vector with complex entries. OK, so be it. It's hard to visualize, perhaps, and doesn't represent a point on the plane, but we can still compute with it just fine. Since lambda_2 = conjugate of lambda_1, an eigenvector for lambda_2 is given by the conjugate of v_1: v_2 = [ 1 ; -i ] So the normal modes _ v(t) = e^{(1+2i)t} [ 1 ; i ] , v(t) = e^{(1-2i)t} [ 1 ; -i ] . As in the case of second order equations, the real and imaginary parts of solutions are again solutions, _ _ u1 = (v + v)/2 = Re(v) , u2 = (v - v)/(2i) and we really only need to write down one of the normal modes. So these are real solutions: u = e^{(1+2i)t} [ 1 ; i ] = e^t ( cos(2t) + i sin(2t) ) ( [1;0] + i[0;1] ) so u1 = Re(u) = e^t ( cos(2t) [1;0] - sin(2t) [0;1] ) = e^t [cos(2t) ; - sin(2t)] and u2 = Im(u) = e^t ( cos(2t) [0;1] + sin(2t) [1;0] )= e^t [sin(2t) ; cos(2t)] These are two independent real solutions. Both spiral around the origin, clockwise, while fleeing away from it exponentially. They satisfy u1(0) = [1;0] , u2(0) = [0;1] . I showed their trajectories on the Mathlet Linear Phase Portraits: Matrix Entry. The general real solution is a u1 + b u2 , a, b real . It is very hard for me to visualize the fact that all those spirals are linear combinations of any two of them. Summary: Nonreal eigenvalues lead to spiral solutions. Positive real parts lead to solutions going to infinity with t ("unstable") Negative real parts lead to solutions going to zero with t ("stable") Zero real parts lead to solutions parametrizing ellipses. So we discover that the possibility of complex eigenvalues really isn't a failure of the method at all. There are in fact ray solutions, but they are complex and don't show up on our real phase plane. [3] Second problem with our method: Illustrated by A = [ -2 1 ; -1 0 ] p_A(lambda) = lambda^2 + 2 lambda + 1 = (lambda + 1)^2 which has only one root, "repeated": lambda_1 = lambda_2 = -1. Still, find an eigenvector: A - (-1)I = [ -1 1 ; -1 1 ][ ? ; ? ] = [ 0 ; 0 ] : v = [ 1 ; 1 ] or any nonzero multiple. ALL eigenvectors for A lie on the line containing 0 and [ 1 ; 1 ]. I showed a picture of the phase portrait, which shows only one pair of opposed ray trajectories. So there is (up to multiples) only one normal mode: u_1 = e^{-t} [ 1 ; 1 ] But we need another solution. Here is how to find one; I won't go into details, just give you the method. Write down the same matrix A - lambda_1 I but now find a vector w such that (A - lambda1 I) w = v . Thenu_2 = e^{lambda_1 t} (t v + w) is a second solution. In our case: [ -1 1 ; -1 1 ] [ ? ; ? ] = [ 1 ; 1 ] has solution [0;1] , so u_2 = e^{-t} ( t [1;1] + [0;1] ) = e^{-t} [ t ; t+1 ] [0;1] isn't the only vector that works here; [0;1] + c v does too for any constant c. It doesn't matter which one you pick. With this choice, u_1(0) = [1;1] , u_2(0) = [0;1] . The general solution is u = a u_1 + b u_2 . To learn more about all this you should take 18.06. Didn't get to talk about this on Friday: [4] A matrix with a repeated eigenvalue but only one lineful of eigenvectors is called "defective." A matrix can have a repeated eigenvalue and not be defective: A = [ 2 0 ; 0 2 ] for example has characteristic polynomial p(lambda) = lambda^2 - 4 lambda + 4 = (lambda - 2)^2 so lambda_1 = lambda_2 = 2. To find …

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