Here is an elegant proof of the identity eA'+B'= eA'eB' if AB-BA. (a) Show that X(t) =
Chapter 3, Problem 17(choose chapter or problem)
Here is an elegant proof of the identity eA'+B'= eA'eB' if AB-BA. (a) Show that X(t) = eAt+B' satisfies the initial-value problem x = (A + B)X, X(0) = I. @) Show that eA'B = BeA' if AB = BA. (Hint: AJB = BAJ if AB = BA). Then, conclude that (d/dt)eA'eBr =(A + B)eA'eBt. (c) It follows immediately from Theorem 4, Section 3.4 that the solution X(t) of the initial-value problem x = (A + B)X, X(0) = I, is unique. Conclude, therefore, that eA'+B' = eA'eB'.
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