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# lecture_36 (dragged) 1 - 2 MA 36600 LECTURE NOTES WEDNESDAY...

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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 A k T = D k for any nonnegative integer k : k = 0: T 1 A 0 T = T 1 I T = I = D 0 k = 1: T 1 A 1 T = D 1 k 2: T 1 A k T = T 1 A k 1 T T 1 A T = T 1 A k 1 T T 1 A T = D k 1 D = D k This gives the expression T 1 exp ( A t ) T = k =0 T 1 A k T t k k ! = k =0 D k t k k ! = exp ( D t ) . The claim follows. Example. Let us review the following homogeneous system considered in the previous lecture: d dt x = 1 1 4 1 x . We found that the general solution is x ( t ) = c 1 1 2 e 3 t + c 2 1 2 e t = Ψ ( t ) c in terms of Ψ ( t ) = e 3 t e t 2 e 3 t 2 e t , c = c 1 c 2 . We explain how to find the fundamental matrix Ψ ( t ) using exponentials of matrices. Denote the 2 × 2 matrices A = 1 1 4 1 and Ψ 0 = 1 1 2 2 . The fundamental matrix above is the unique solution to the initial value problem d dt Ψ = A Ψ , Ψ (0) = Ψ 0 . We know that this has the unique solution
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