3.2-3.4z

3.2-3.4z - a transformation T : F m F n such that L A T is...

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1 Math 110 Homework 8 Partial Solutions If you have any questions about these solutions, or about any problem not solved, please ask via email or in office hours, etc. 3.2.17 Suppose that B = a b c and C = ± x y z ² . Then A = ax ay az bx by bz cx cy cz surely has rank at most 1, since its row space is spanned by the one vector ( x, y, z ). On the other hand, suppose that A has rank 1. Then the row space of A is spanned by one non-zero vector, say ( x, y, z ). In this case, every row is a multiple of ( x, y, z ). So A = ax ay az bx by bz cx cy cz for some a, b, c F . But then A = BC as above. 3.2.19 Given such A and B , we have that L A is a linear transformation from F n to F m , and that L B is a linear transformation from F p to F n . Since rank( L A ) = rank( A ) = m and rank( L B ) = rank( B ) = n , we have that L A and L B are onto. Thus L AB = L A L B is onto and has rank m . Thus rank( AB ) = m . 3.2.21 Since A is m × n of rank m , L A maps from F n onto F m . We construct
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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.

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