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Unformatted text preview: Honours Algebra 2, Assignment 1 Michael Snarski & Jamie Klassen January 20, 2012 Question 1. Show that the set V = R 2 , with the addition and scalar multi plication defined by the rules ( x 1 ,y 1 ) + ( x 2 ,y 2 ) := ( x 1 + x 2 ,y 1 + y 2 ) , ( x,y ) := ( x, 0) satisfies all the axioms of a vector space except the property that 1 v = v for all v V . Solution This will eventually become an exercise in calligraphy (or typing, if you have already graduated to L A T E X), but it is important to do these so that you realize there is actually something to prove. i. Certainly 0 V V , as (0 , 0) satisfies the properties of a zero vector: (0 , 0) + ( x,y ) = ( x,y ) + (0 , 0) = ( x + 0 ,y + 0) = (0 + x, 0 + y ) = ( x,y ). ii. For any v V , v V . Given v = ( x,y ), take v = ( x, y ). Note that v negationslash = 1 v , which is ( x, 0) by definition of scalar multiplication. We define the additive inverse v to be ( x, y ). On the other hand, in an actual vector space, v = 1 v (this fails here). iii. Associativity of addition (ugh); we must show that ( v + w )+ z = v +( w + z ) for all v,w,z V . This follows from associativity in the field, but we can simply note that addition is defined just as in R 2 (componentwise), which we know to be a vector space; hence associativity follows. iv. If you wanted a proper answer for iii. , heres the standard version for commutativity of addition: for any v 1 = ( x 1 ,y 1 ) ,v 2 = ( x 2 ,y 2 ) V , v 1 + v 2 = ( x 1 ,y 1 ) + ( x 2 ,y 2 ) = ( x 1 + x 2 ,y 1 + y 2 ) = ( x 2 + x 1 ,y 2 + y 1 ) = ( x 2 ,y 2 ) + ( x 1 ,y 1 ) = v 2 + v 1 v. For any , R , v V , ( ) v = ( ) ( x,y ) = ( x, 0) = ( x, 0) = ( ( x, 0)) = ( ( x,y )) = ( v ) which is the associativity law for scalar multiplication. vi. Distributivity of scalars over vectors: for any , R , v V , ( + ) v = ( + ) ( x,y ) = (( + ) x, 0) = ( x + x, 0) = ( x, 0) + ( x, 0) = ( x, 0) + ( x, 0) = ( x,y ) + ( x,y ) = v + v 1 vii. Distributivity of vectors over scalars: for any R , v 1 ,v 2 V , ( v 1 + v 2 ) = (( x 1 ,y 1 ) + ( x 2 ,y 2 )) = ( x 1 + x 2 ,y 1 + y 2 ) = ( ( x 1 + x 2 ) , 0) = ( x 1 + x 2 , 0) = ( x 1 , 0) + ( x 2 , 0) = ( x 1 , 0) + ( x 2 , 0) = ( x 1 ,y 1 ) + ( x 2 ,y 2 ) = v 1 + v 2 viii. Indeed, the last one, which doesnt hold, is that the identity acts triv ially on any element v V : 1 R ( x,y ) = ( x, 0) negationslash = ( x,y ). [Some gra tuitous knowledge/remark: the word acts here is appropriate; scalar multiplication defines a group action on the set (commutative group) ( V, +) when V is a vector space. In this case, it is the requirement for R V V to define an action, 1 R v = v for all v V , which fails.] The only nonstandard thing is when we write...
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This note was uploaded on 01/25/2012 for the course PHIL 201H1F taught by Professor Derekallen during the Fall '10 term at University of Toronto Toronto.
 Fall '10
 DerekAllen
 Philosophy

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