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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 22 Now we return to the general case. Consider an n × n matrix A. Say that there exists a nonsingular n × n matrix T such that T−1 A T = D is an n × n diagonal matrix. Such a matrix A is said to be diagonalizable. We will show that exp (A t) = T exp (D t) T−1 . In practice this is easy to compute because both T−1 and exp (D t) are easy to compute. Observe that T−1 Ak T = Dk for any nonnegative integer k : k = 0: T−1 A0 T = T−1 I T = I = D0 k = 1: T−1 A1 T = D1 ￿ ￿ ￿ ￿￿ ￿ k ≥ 2: T−1 Ak T = T−1 Ak−1 T T−1 A T = T−1 Ak−1 T T−1 A T = Dk−1 D = Dk This gives the expression T−1 exp (A t) T = ∞ ￿￿ k=0 The claim follows. ∞ ￿ ￿ tk tk T−1 Ak T = Dk = exp (D t) . k! k! k=0 Example. Let us review the following homogeneous system considered in the previous lecture: ￿ ￿ d 11 x= x. 41 dt We found that the general solution is ￿￿ ￿￿ ￿ 3t ￿ ￿￿ 1 3t 1 −t e e−t c x(t) = c1 e + c2 e = Ψ(t) c in terms of Ψ(t) = , c= 1 . 2 −2 2 e3t −2 e−t c2 We explain how to find the fundamental matrix Ψ(t) using exponentials of matrices. Denote the 2 × 2 matrices ￿ ￿ ￿ ￿ 11 1 1 and Ψ0 = . A= 41 2 −2 The fundamental matrix above is the unique solution to the initial value problem d Ψ = A Ψ, Ψ(0) = Ψ0 . dt We know that this has the unique solution Ψ(t) = exp (A t) Ψ0 . In order to exponentiate the matrix A, we diagonalize the matrix. Consider the matrices ￿ ￿ ￿ ￿ 3 0 1 1 D= and T = Ψ0 = . 0 −1 2 −2 Then we have the product ￿ ￿￿ 1 −2 −1 1 −1 T AT = 14 −4 −2 Hence we have the matrix exponential 1 1 ￿￿ 1 2 ￿ ￿ 1 −2 1 = −2 −4 −2 ￿￿ −1 3 16 exp (A t) = T exp (D t) T−1 ￿ ￿￿ ￿ ￿ ￿￿ 1 1 e3t 0 1 −2 −1 1 = = 2 −2 0 e−t −4 −2 1 2 ￿ 1 3t 1 −t 1 3t 1 −t ￿ e+e 4e −4e = 2 3t 2−t 1 3t 1 −t . e −e 2e +2e This gives the fundamental matrix ￿ 1 Ψ(t) = exp (A t) T = T exp (D t) = 2 1 −2 ￿ ￿ 3t e 0 ￿￿ −1 3 = 2 0 1 −2 ￿￿ 1 ￿ ￿ 0 e−t 3t 2e 1 −t 2e e3t = 2 e3t ￿ 0 = D. −1 1 3t 4e 1 −t −4 e ￿ ￿ e−t . −2 e−t ...
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