# sol01 - Honours Algebra 2 Assignment 1 Michael Snarski&...

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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 ) ,λ · ( x,y ) := ( λx, 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. , here’s 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 α,β ∈ R , v ∈ V , ( αβ ) · v = ( αβ ) · ( x,y ) = ( αβx, 0) = α · ( βx, 0) = α · ( β · ( x, 0)) = α · ( β · ( x,y )) = α · ( β · v ) which is the ‘associativity’ law for scalar multiplication. vi. Distributivity of scalars over vectors: for any α,β ∈ R , v ∈ V , ( α + β ) · v = ( α + β ) · ( x,y ) = (( α + β ) x, 0) = ( αx + βx, 0) = ( αx, 0) + ( βx, 0) = α · ( x, 0) + β · ( x, 0) = α · ( x,y ) + β · ( x,y ) = αv + βv 1 vii. Distributivity of vectors over scalars: for any α ∈ R , v 1 ,v 2 ∈ V , α · ( v 1 + v 2 ) = α · (( x 1 ,y 1 ) + ( x 2 ,y 2 )) = α · ( x 1 + x 2 ,y 1 + y 2 ) = ( α ( x 1 + x 2 ) , 0) = ( αx 1 + αx 2 , 0) = ( αx 1 , 0) + ( αx 2 , 0) = α · ( x 1 , 0) + α · ( x 2 , 0) = α · ( x 1 ,y 1 ) + α · ( x 2 ,y 2 ) = α · v 1 + α · v 2 viii. Indeed, the last one, which doesn’t 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 non-‘standard’ thing is when we write...
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sol01 - Honours Algebra 2 Assignment 1 Michael Snarski&...

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