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204PS4Solution2009

# 204PS4Solution2009 - Economics 204 Problem Set 4 Solutions...

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Economics 204 Problem Set 4 Solutions Exercise 1 a) First note that S is a subset of R 3 ; which is a vector space (over R ; with the operations assumed). Hence, all we have to show is that 0 vector is contained in S and that 8 °; ± 2 R , and x; y 2 S , we have °x + ±y 2 S . But this is pretty obvious: i ) take c = 0 to show that 0 2 S ; ii ) if x = c 1 v and y = c 2 v , then °x + ±y = ( °c 1 + ±c 2 ) v and if we let c = °c 1 + ±c 2 then it follows that °x + ±y 2 S: The space is one dimensional, and f v g is a basis for S . b) Same argument applies here: i ) 0 vector is obviously in S . ii ) Now take °; ± 2 R and x; y 2 S ; let z := °x + ±y , then z 1 + z 2 + z 3 = ( °x 1 + ±y 1 ) + ( °x 2 + ±y 2 ) + ( °x 3 + ±y 3 ) = ° ( x 1 + x 2 + x 3 ) + ± ( y 1 + y 2 + y 3 ) = ° 0 + ± 0 = 0; and z 1 + 2 z 2 = ( °x 1 + ±y 1 ) + 2( °x 2 + ±y 2 ) = ( °x 1 + 2 °x 2 ) + ( °y 1 + 2 ±y 2 ) = ° ( x 1 + 2 x 2 ) + ± ( y 1 + 2 y 2 ) = 0 . The space is again one dimensional (Note that if we °x x 2 , then x 1 and x 3 are determined). f (1 ; ° 1 ; 0) g is a basis for S . c) S is not a vector space since it does not contain 0 vector. d) not a vector space since not all additive inverses of continuous functions are in S: Exercise 2 a) x 2 Ker ( g ) = > g ( x ) = 0 = > ( f ± g )( x ) = f ( g ( x )) =

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