Advanced Linear Algebra Questions Let V denote the set R2 to

Advanced Linear Algebra Questions

Let V denote the set R^2 together with the operations (a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2b_2), alpha(a_1, a_2) = (alpha a_1, a_2) Is V a vector space over R with these operations? Justify your answer.

Solution

let a=(a1,a2) b=(b1,b2) c=(c1,c2)

1. here a+b=b+a as (a1+b1,a2b2)=(b1+a1,b2a2)

2.now a+(b+c)=(a1,a2)+(b1+c1,b2c2)=(a1+b1+c1,a2b2c2)

      and (a+b)+c=(a1+b1,a2b2)+(c1,c2)=(a1+b1+c1,a2b2c2)

so a+(b+c)=(a+b)+c

3.(a1,a2)+(0,1)=(a1+0,a2*1)=(a1,a2)   hence (0,1) is the null element which exists in V

4.(a1,a2)+(-a1,1/a2)=(a1-a1,a2*1/a2)=(0,1)   so the inverse element exists in V

5. k(p*a)=k(pa1,a2)=(kpa1,a2)

   p(k*a)=p(ka1,a2)=(kpa1,a2)

hence k(p*a)= p(k*a)

6.k(a+b)=k(a1+b1,a2b2)=(ka1+kb1,a2b2)

and (ka)+(kb)=(ka1,a2)+(kb1,b2)=(ka1+kb1,a2b2)

so k(a+b)=(ka)+(kb)

7. 1*a=(a1,a2) so 1 is the identity element

hence it satisfies all the conditions.

hence V is a vector space. [proved]