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lecture_33 (dragged) 2 - MA 36600 LECTURE NOTES WEDNESDAY...

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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 15 3 The characteristic polynomial is ￿ ￿ ￿ ￿ b c a r2 + b r + c r −1 pA (r) = det =r r+ − (−1) · = . c/a r + b/a a a a Hence an eigenvalue r of A is a root of the characteristic equation a r2 + b r + c = 0. More generally, consider the constant coeﬃcient equation dn y dn−1 y dy + a1 n−1 + · · · + an−1 + an y = 0. n dt dt dt We found earlier in the lecture that if we substitute 0 1 0 ··· y 0 1 ··· 0 y (1) . . . .. d . . . . . . . . x= =⇒ x= . . dt 0 y (n−2) 0 0 ··· a an−1 an−2 n − − − ··· y (n−1) a0 a0 a0 ￿ ￿￿ a0 0 0 0 0 . . . . . . 0 a2 − a0 1 a1 − a0 A x. ￿ The characteristic polynomial is Z (r) pA (r) = det (r I − A) = in terms of Z (r) = a0 rn + a1 rn−1 + · · · + an−1 r + an . a0 Hence an eigenvalue of A is a root of the characteristic equation Z (r) = 0. 2 × 2 Matrices. We focus on the case where A is a 2 × 2 constant matrix: ￿ ￿ a a12 A = 11 . a21 a22 Say that we can ﬁnd two functions x(1) (t) and x(2) (t) such that i. x(k) (t) is a solution to x￿ = A x for k = 1, 2; ii. their Wronskian is nonzero: ￿ ￿ ￿x11 (t) x12 (t)￿ ￿ (1) (2) ￿ ￿ ￿ W x , x (t) = ￿ where x21 (t) x22 (t)￿ x (k ) Then the general solution to x￿ = A x is ￿ ￿ x1k (t) (t) = . x2k (t) x(t) = c1 x(1) (t) + c2 x(2) (t) for constants c1 and c2 . We can ﬁnd such functions x(k) (t) by making a guess in the form x(t) = ξ ert . We focus on the eigenvalues r of A. The characteristic polynomial is pA (r) = det (r I − A) ￿ ￿ r − a11 −a12 = det = (r − a11 ) (r − a22 ) − a12 a22 = r2 − (a11 + a22 ) r + (a11 a22 − a12 a21 ) . −a21 r − a22 Recall that the trace and determinant of A are the quantities tr A = a11 + a22 and det A = a11 a22 − a12 a21 . To recap, we have shown that the characteristic polynomial of a 2 × 2 matrix is pA (r) = r2 − (tr A) r + (det A) . Note that the discriminant of this polynomial is 2 2 disc A = (tr A) − 4 (det A) = (a11 − a22 ) + 4 a12 a21 . If the discriminant is positive, then the eigenvalues r are real. If the discriminant is negative, then the eigenvalues r are imaginary. ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

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