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235_tt1_f10_soln - Math 235 1 Short Answer Problems Term...

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Math 235 Term Test 1 Solutions 1. Short Answer Problems [1] a) State the definition of the rank of a linear mapping L : V W . Solution: rank( L ) = dim Range( L ). [2] b) Let B = { v 1 , . . . , v n } be a basis for a vector space V and let L : V W be an isomorphism. Determine a basis for W . Solution: { L ( v 1 ) , . . . , L ( v n ) } is a basis for W . [2] c) Let A = 1 3 - 1 2 . Determine A under the inner product A, B = tr( B T A ) for M (2 , 2). Solution: A = A, A = 15 [2] d) Determine if p, q = p ( - 1) q ( - 1) + p (1) q (1) is an inner product for P 2 . Solution: It is not an inner product since x 2 - 1 , x 2 - 1 = 0 but x 2 - 1 is not the zero vector in P 2 . [3] e) If P is orthogonal and symmetric, prove that 1 2 ( I - P ) = 1 2 ( I - P ) 2 Solution: 1 2 ( I - P ) 2 = 1 4 [ I 2 - P - P + P 2 ] = 1 4 [ I - 2 P + P T P ] since P = P T = 1 4 [ I - 2 P + I ] since P T P = I = 1 4 [2 I - 2 P ] = 1 2 [ I - P ] 1
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2. Consider the linear transformation L : M (2 , 2) P 3 defined by L a b c d = ( a + b ) x 3 + cx [2] a) Find a basis B = { v 1 , v 2 , v 3 , v 4 } of M (2 , 2) such that the first k vectors, { v 1 , · · · , v k } , in the basis B forms a basis for Null( L ). Solution: A basis for the nullspace is for example: 1 - 1 0 0 , 0 0 0 1 (There are many correct answers, but all consist of two elements, i.e. k = 2.) Extend this to a basis of M (2 , 2). For example: 1 - 1 0 0 , 0 0 0 1 , 1 0 0 0 , 0 0 1 0 [1] b) Is { L ( v k +1 ) , ..., L ( v n ) } a basis for the range of L ? Solution: This is a basis of the range of L , by the argument in the proof of the Rank-Nullity Theorem. Alternatively we can check it directly: L 1 0 0 0 = x 3 and L 0 0 1 0 = x [2] c) State the rank and nullity of L and verify the Rank-Nullity Theorem. Solution: The Rank-Nullity theorem says for a linear mapping L : V W we have rank( L ) + nullity( L ) = dim V .
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