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Exercise 5.3.24

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Use the notation in the solution to Exercise 5.2.15. And in the proof we know that the solution $x (t) = Q \bar{D} y$ for some $y \in ℝ^{n}$ . Now we know that actually $\bar{D} = e^{D}$ by definition. With the fact that $Q$ is invertible, we may write the set of solution to be

{Q e^{D} Q^{- 1} v : v \in ℝ^{n}} = {e^{A} v : v \in ℝ^{n}}

by Exercise 5.3.21.

jephianlin

2011-06-27 00:00

Exercise 5.3.24

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