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2.42.5 - 2.4 Problem 7 Let A be an n n matrix Theorem If A2...

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§ 2.4, Problem 7 Let A be an n × n matrix. Theorem. If A 2 = 0 , then A is not invertible Proof. For a contradiction assume that A 2 = 0 and A is invertible. Then A has an inverse A - 1 and A 2 = 0 , A - 1 AA = A - 1 0 , IA = 0 , A = 0 . But then A - 1 A = A - 1 0 = 0 = I – Contradiction. Theorem. If AB = 0 for some nonzero n × n matrix B , then A is not invertible. Proof. Again, by contradiction. Assume AB = 0 and A is invertible. Then AB = 0 , A - 1 AB = A - 1 0 , IB = 0 , B = 0 . Contradiction: we assumed B was nonzero. § 2.4, Problem 15 Theorem. Let V and W be finite-dimensional vector spaces, and let T : V W be a linear transformation. Suppose that β = { β 1 , . . . , β n } is a basis for V . Then T is an isomorphism if and only if T ( β ) is a basis for W . Proof. ( ) First assume that T is an isomorphism. Since T an isomorphism it is one-to-one and onto. Problem 14(c) on page 75 applies to tell us that T ( β ) is a basis. ( ) Conversely, assume that T ( β ) is a basis for W . (We want to prove that T is an isomorphism. Since we already know T is linear we need prove that T is one-to-one and onto.)
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