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Unformatted text preview: Linear Algebra 115 Solutions to Seventh Homework Problem 13 (Section 2 . 2) Assume that T,U are not linearly independent. Then there are some a,b not both zero such that aT + bU = 0. By assumption we know that T is not the zero transformation, so there is a x ∈ V with T ( x ) 6 = 0. For this x we have T ( ax ) = aT ( x ) = bU ( x ) = U ( bx ). But T ( ax ) ∈ R ( T ) and U ( bx ) ∈ R ( U ). Since their intersection contains only the zero vector we have that T ( ax ) = 0, or that aT ( x ) = 0 with T ( x ) 6 = 0. Thus, a = 0. Then bU = 0 and U is a nonzero transformation. So, it must be b = 0. Combining these two, a = b = 0 we get a contradiction. So, U,T are linearly independent. Problem 15 (Section 2 . 2) Let β = { v 1 ,v 2 ,...,v k } be a basis for N ( T ), with dim ( N ( T )) = k ≤ n = dim ( V ). By Replacement Theorem we know that we can extend it to a basis β = { u 1 ,...,u n k ,v 1 ,v 2 ,...,v k , } for V . We claim that { T ( u 1 ) ,T ( u 2 ) ,...,T ( u n k ) } is a linearly independent subset of W . Let a 1 T ( u 1 ) + ... + a n T ( u n k ) = ⇒...
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 Winter '07
 Liu
 Linear Algebra, Algebra, basis, linearly independent subset, 0wk, Replacement Theorem

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