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# Linear Alegbra : Vector Space

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Let V= (x,y) in R2{y=3x+1} with addition and multiplication by a scalar defined on V by:
(x,y)+ (x',y')= (x+x',y+y'-1)
k(x,y)=(kx,k(y-1)+1)

Given that with these definitions, V satisfies vector space axioms 1,2,3,6,8,9,and 10 determine whether or not V is a vector space by checking to see if axioms 4,5,7,are also satisfied.

##### Solution Summary

A vector space is defined.

##### Solution Preview

We have:

axiom 4:
There exists an element namely 0 in V such that for each v in V we have: v+0=v.
Ok, as V is in R2, we consider a sample element like v to be of form(x,y), we also consider 0 in V to have the form (x',y') and now we have to find x' and y' if there exist any such that:
(x+y)+(x',y')=(x,y)
(x+x',y+y'-1)=(x,y) ---> x+x'=x, y+y'-1=y ---> ...

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