MAS4105Su10b

# MAS4105Su10b - MAS 4105 Test 2 1(8 pts In the following let...

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Unformatted text preview: MAS 4105 Test 2 1. (8 pts) In the following let V , W , and Z be vector spaces of finite dimension over a field F with ordered bases β, γ, and δ, respectively, and let T : V → W and U : W → Z be linear transformations. Indicate whether the following are true or false. i. If N ( T ) = { v } , then T is one-to-one. T F ii. If U is one-to-one and T is an isomorphism, then nullity( UT ) = 0. T F iii. W is isomorphic to Z if and only if dim( W ) = dim( Z ). T F iv. If U and T are invertible then UT is invertible. T F v. If dim( W ) = dim( Z ), then U is onto. T F vi. If v ∈ V , then [ UT (v)] δ = [ U ] δ γ [ T ] γ β [v] β . T F vii. If T is invertible, then ([ T ] γ β )- 1 = ([ T- 1 ] γ β ). T F viii. If dim( W ) = dim( Z ) and U ( γ ) = δ , then U is invertible. T F 2. (5 pts) Explain how the underlined statement is equivalent to Φ being one-to-one and onto. Theorem 2.20. Let V and W be finite-dimensional vector spaces over F of dimensions n and m , respectively, and let β and γ be ordered bases for V and W respectively. Then the function Φ : L ( V , W ) → M m × n ( F ), defined by Φ ( T ) = [ T ] γ β for T ∈ L ( V , W ) is an isomorphism. Proof. By Theorem 2.8, Φ is linear. Hence we must show that Φ is one-to-one and onto. This is accomplished if we show that for every m × n matrix A , there exists a unique linear transformation T : V → W such that Φ ( T ) = A. Let β = { v 1 , v 2 , . . . , v n } , γ = { w 1 , w 2 , . . . , w n } , and A be a given m × n matrix. By Theorem 2.6, there exists a unique linear transformation T : V → W such that T (v j ) =...
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MAS4105Su10b - MAS 4105 Test 2 1(8 pts In the following let...

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