MIT18_03S10_c34

MIT18_03S10_c34 - 18.03 Class 34 Complex or repeated...

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, 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]

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This note was uploaded on 01/18/2012 for the course MATH 18.03 taught by Professor Vogan during the Spring '09 term at MIT.

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MIT18_03S10_c34 - 18.03 Class 34 Complex or repeated...

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