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Linear Algebra

Exercise 2.7.15

Answers

1.: The mapping $Φ$ is linear since the differential operator $D$ is linear. If $Φ (x) = 0$ , then $x$ is the zero function by the previous exercise. Hence $Φ$ is injective. And the solution space is an $n$ -dimensional space by Theorem 2.32. So The mapping is an isomorphism.
2.: This comes from the fact the transformation $Φ$ defined in the previous question is an isomorphism.

jephianlin

2011-06-27 00:00

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