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Unformatted text preview: 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 k A k under the inner product h A, B i = tr( B T A ) for M (2 , 2). Solution: k A k = p h A, A i = 15 [2] d) Determine if h p, q i = p ( 1) q ( 1) + p (1) q (1) is an inner product for P 2 . Solution: It is not an inner product since h x 2 1 , x 2 1 i = 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 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 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 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 RankNullity 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 RankNullity Theorem....
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.
 Spring '10
 WILKIE
 Vector Space

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