Ma1bAnHw3Sol

# Ma1bAnHw3Sol - CALIFORNIA INSTITUTE OF TECHNOLOGY...

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CALIFORNIA INSTITUTE OF TECHNOLOGY Department of Mathematics Math 1b; Solutions to Homework Set 3 Due: January 28, 2008 1. (a) ( ST )( x,y,z ) = S ( T ( x,y,z )) = S ( x,x + y,x + y + z ) = ( x + y + z,x + y,x ). Similarly ( TS )( x,y,z ) = T ( z,y,x ) = ( z,y + z,x + y + z ). Hence ( ST - TS )( x,y,z ) = ( ST )( x,y,z ) - ( TS )( x,y,z ) = ( x + y + z,x + y,x ) - ( z,y + z,x + y + z ) = ( x + y,x - z, - y - z ) . Next S 2 = I , T 2 ( x,y,z ) = ( x, 2 x + y, 3 x + 2 y + z ), ( ST ) 2 ( x,y,z ) = (3 x + 2 y + z, 2 x + 2 y + z,x + y + z ), and ( TS ) 2 ( x,y,z ) = ( x + y + z,x + 2 y + 2 z,x + 2 y + 3 z ). ( ST - TS ) 2 ( x,y,z ) = ( ST - TS )( x + y,x - z, - y - z ) = (2 x + y - z,x +2 y + z, 2 z + y - x ) . (b) From (a), S 2 = I , so S = S - 1 , and hence by Lemma 1B, S is a 1-1 correspondence as it has an inverse. Let g ( x,y,z ) = ( x,y - x,z - y ). Then T ( g ( x,y,z )) = T ( x,y - x,z - y ) = ( x,y,z ), so T g = I . Similarly g T = I , so g = T - 1 and T is a 1-1- correspondence. Finally ( ST

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## This note was uploaded on 04/28/2010 for the course MATH 1B taught by Professor Aschbacher during the Winter '07 term at Caltech.

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Ma1bAnHw3Sol - CALIFORNIA INSTITUTE OF TECHNOLOGY...

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