lecture_33 (dragged) 2

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

Ask a homework question - tutors are online