let v be the vector space of all ordered pair of real number check weather v is a vector space over t with respect to indicated operations if not state the axioms which fail to hold(a,b)+(c,d)=(a+b,c+d) K(a,b)=(K^2a,K^2b)
"k((a_1,b_1),(a_2,b_2))=k(a_1,b_1)+k(a_2,b_2)=k^2(a_1,b_1)+k^2(a_2,b_2)"
So, axiom [M1] does not hold. That is, this is not a vector space.
Comments
Leave a comment