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Unformatted text preview: a transformation T : F m F n such that L A T is the identity. Then T = L B for some matrix B , and this is the matrix we seek as then I m = [ L A T ] = [ L A L B ] = [ L AB ] = AB . We construct T by prescribing its behavior on a basis for F m . Let { e 1 , . . . , e m } be the standard basis for F m . For each e i , there exists x i F n such that L A ( x i ) = e i since L A is onto. Dene T by setting T ( e i ) = x i for each i . Then L A ( T ( e i )) = e i for every i , and so L A T is the identity on F m . This proves the claim and the result. 3.3.10 This statement is true. If the m n coecient matrix A has rank m , then L A : F n F m is onto. Thus (by previous hw problem) any equation Ax = b has a solution....
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
 Winter '07
 Liu
 Math, Linear Algebra, Algebra

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