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Unformatted text preview: Winter 2011 • Math 67 • Linear Algebra Homework 4 Problems 3.3, 3.4, 3.5, 3.7 and 3.9 from Axler. Solutions. 3.3 Let U ⊂ V be a subspace and S : U → W be a linear map. Pick a basis ( u 1 ,...,u ‘ ) of U and extend it to a basis of V , ( u 1 ,...,u ‘ ,v ‘ +1 ,...,v n ) . Define a map T : V → W by T ( u i ) = S ( u i ) and T ( v j ) = W and extending by linearity. If u ∈ U then write u = ∑ i c i u i and see that T ( u ) = ∑ i c i S ( u i ) = S ( u ), as we needed to show. 3.4 If T is the zero map then null( T ) = V and the question doesn’t really make sense. Assume that T is not the zero map and pick some element u that is not mapped to zero. The intersection null( T ) with { au : a ∈ F } is zero, since u / ∈ null( T ). We apply the rank nullity theorem to see that dim(null( T )) = dim( V ) rank( T ). Any nonzero linear map V → F is surjective, since any non zero element of F is a scalar multiple of another. Thus dim(null( T )) = dim( V ) 1. This means V = null( T ) + { au : a ∈ F } , since the sum is at least n dimensional. 3.5 Suppose that ( v 1 ,...,v n ) is linearly independent. If ( T ( v 1 ) ,...,T ( v n )) were linear dependent there would be scalars a i , not all zero, such that ( * ) X i a i T ( v i ) = = ⇒ T ( X i a i v i ) = That is, ∑ i a i v i is in the null space of T . But if T is injective, the null space is zero. This means ∑ i a i v i...
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This note was uploaded on 11/13/2011 for the course MATH 67 taught by Professor Schilling during the Winter '07 term at UC Davis.
 Winter '07
 Schilling
 Linear Algebra, Algebra

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