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lecture_37 (dragged) 1 - 2 MA 36600 LECTURE NOTES: FRIDAY,...

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Unformatted text preview: 2 MA 36600 LECTURE NOTES: FRIDAY, APRIL 23 Example. Again, consider the 2 × 2 matrix ￿ 1 A= 4 ￿ 1 1 =⇒ tr A = −2 det A = −3 =⇒ disc A = 16 pA (r) = r2 − 2 r − 3 = (r − 3) (r + 1) . Hence the eigenvalues are r1 = 3 and r2 = −1, with corresponding eigenvectors ￿ ￿ ￿￿ ￿ ￿￿￿ a12 1 a12 1 (1) (2) ξ= = and ξ= = . r1 − a11 2 r2 − a11 −2 This gives the matrices D= ￿ r1 0 ￿￿ 0 3 = r2 0 0 −1 ￿ and T= ￿ ξ11 ξ21 ￿￿ ξ12 1 = ξ22 2 ￿ 1 . −2 Repeated Eigenvalues Set-Up. So far we have considered the case of 2 × 2 matrices with distinct eigenvalues. We now consider the case where disc A = 0. To gain some intuition, consider the following second order differential equation with constant coefficients: a y ￿￿ + b y ￿ + c y = 0. Say that the characteristic equation a r2 + b r + c = 0 has repeated roots i.e., b − 4 a c = 0. The roots are r1 = r2 = −b/(2 a) so that the general solution of the differential equation is 2 y (t) = c1 y1 (t) + c2 y2 (t) in terms of y1 (t) = er1 t , y2 (t) = t er2 t . Recall that we can express this differential equation as a system of first order differential equations: ￿￿ ￿ ￿ d 0 1 y x= ￿ =⇒ x = Ax in terms of A= c b. y −a −a dt ￿ ￿ We see that disc A = b2 − 4 a c /a2 = 0. The general solution to this system is ￿￿ ￿ ￿ 1 r1 t t (1) (2) (1) (2) x(t) = c1 x (t) + c2 x (t) in terms of x (t) = e, x (t) = er1 t . r1 r1 t + 1 We see that the general solution cannot be found by writing x(t) = ξ ert for some constant vector ξ and some constant scalar r. Example. Consider the following 2 × 2 matrix: We have the characteristic polynomial ￿ r−1 pA (r) = det (r I − A) = det −1 ￿ 1 A= 1 ￿ −1 . 3 ￿ 1 2 = (r − 1) (r − 3) + 1 = r2 − 4 r + 4 = (r − 2) . r−3 Hence we have equal eigenvalues r = 2. To compute the eigenvectors, we set ￿￿ ￿￿ ￿ ￿￿ ￿ ￿ ￿ ξ 0 1 1 ξ1 ξ1 + ξ2 = (r I − A) ξ = = ξ= 1 =⇒ ξ2 0 −1 −1 ξ2 −1 (ξ1 + ξ2 ) so that ξ2 = −ξ1 . Up to scalar multiple, r=2 has the eigenvector ￿ ￿ 1 ξ= . −1 ...
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