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Unformatted text preview: EE221A Problem Set 3 Solutions  Fall 2011 Problem 1. a) A w.r.t. the standard basis is, by inspection, A E = 2 1 4 2 . b) Now consider the diagram from LN3, p.8. We are dealing with exactly this situation; we have one matrix representation, and two bases, but we are using them in both the domain and the codomain so we have all the ingredients. So the matrices P and Q for the similarity transform in this case are, P = e 1 e 2 e 3 1 b 1 b 2 b 3 = b 1 b 2 b 3 , since the matrix formed from the E basis vectors is just the identity; and Q = b 1 b 2 b 3 1 e 1 e 2 e 3 = b 1 b 2 b 3 1 = P 1 . Let A B be the matrix representation of A w.r.t. B . From the diagram, we have A B = QA E P = P 1 A E P = b 1 b 2 b 3 1 A E b 1 b 2 b 3 = 1 2 5 2 1 1  1 2 1 4 2 1 2 5 2 1 1 = 1 15 16 4 12 7 32 6 21 6 12 Problem 2. Representation of a linear map. This is straightforward from the definition of matrix represen tation, A = 1 ....
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 Fall '10
 ClaireTomlin

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