Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 15 3 The characteristic polynomial is
a r2 + b r + c
pA (r) = det
− (−1) · =
c/a r + b/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
+ a1 n−1 + · · · + an−1
+ an y = 0.
We found earlier in the lecture that if we substitute 0
0 y (1) . .
. dt 0 y (n−2) 0
a0 0 0 0 0 .
a0 A x. The characteristic polynomial is
pA (r) = det (r I − A) =
in terms of
Z (r) = a0 rn + a1 rn−1 + · · · + an−1 r + an .
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 = 11
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)
W x , x (t) =
x21 (t) x22 (t) x (k ) Then the general solution to x = A x is
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
= (r − a11 ) (r − a22 ) − a12 a22 = r2 − (a11 + a22 ) r + (a11 a22 − a12 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.
- Spring '09