MIT18_03S10_ls3

MIT18_03S10_ls3 - LS.3 Complex and Repeated Eigenvalues 1...

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± ± ± LS.3 Complex and Repeated Eigenvalues 1. Complex eigenvalues. In the previous chapter, we obtained the solutions to a homogeneous linear system with constant coefficients A x = 0 under the assumption that the roots of its characteristic equation | A I | = 0 i.e., the eigenvalues of A were real and distinct . In this section we consider what to do if there are complex eigenvalues. Since the chan- racteristic equation has real coefficients, its complex roots must occur in conjugate pairs: ¯ = a + bi, = a bi . Let’s start with the eigenvalue a + bi . According to the solution method described in Chapter LS.2, the next step would be to Fnd the corresponding eigenvector λ± , by solving the equations (LS.2, (27)) ( a ) a 1 + ba 2 = 0 ca 1 + ( d ) a 2 = 0 for its components a 1 and a 2 . Since is complex, the a i will also be complex, and therefore the eigenvector λ± corresponding to will have complex components. Putting together the eigenvalue and eigenvector gives us formally the complex solution λ (1) x = ± e ( a + bi ) t . Naturally, we want real solutions to the system, since it was real to start with. To get them, the following theorem tells us to just take the real and imaginary parts of (1). Theorem 3.1 Given a system x = A x , where A is a real matrix. If x = x 1 + i x 2 is a complex solution, then its real and imaginary parts x 1 , x 2 are also solutions to the system. Proof. Since x 1 + i x 2 is a solution, we have ( x + i x 2 ) = A ( x + i x 2 ) ; equating real and imaginary parts of this equation, 1 = A x 1 , x 2 = A x 2 , x which shows that the real vectors x 1 and x 2 are solutions to x = A x . Example 1. ±ind the corresponding two real solutions to x = A x if a complex eigenvalue and corresponding eigenvector are i λ = 1 + 2 i , ± = . 2 2 i Solution. ±irst write λ± in terms of its real and imaginary parts: 0 1 λ± = + i . 2 2 13
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14 18.03 NOTES: LS. LINEAR SYSTEMS The corresponding complex solution x = λ e t to the system can then be written ± ±± 1 ² ³ 0 x = + i e t cos 2 t + i sin 2 t , 2 2 so that we get respectively for the real and imaginary parts of x ± ± ± ± 0 1 sin 2 t t x 1 = e cos 2 t i sin 2 t = e t 2 2
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MIT18_03S10_ls3 - LS.3 Complex and Repeated Eigenvalues 1...

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