Determine whether the given set, together with the specified operations of addition and scalar multiplication, is a vector space. If it is not, select all of the axioms that fail hold. (Let u, v, and w be vectors in the vector space V, and let c and d be scalars.)R2, with the usual addition but scalar multiplication defined by{]-[*}All of the axioms hold, so the given set is a vector space.1. u + v is in V.U2. u + V = V + U3. (u + v) + w = u + (v + w)4. There exists an element 0 in V, called a zero vector, such that u + 0 = u.5. For each u in V, there is an element -u in V such that u + (-u) = 0.6. cu is in V.O 7. c(u + v) = cu + cv8. (c + d)u = cu + du9. c(du) = (cd)u10. 1u = USnipping Tool

Given: u, v, and w are vectors in the vector space V and c and d are scalars in ℝ2 with usual addition and the scalar multiplication is defined by:cxy=cxy We have to determine whether the given set, with addition and scalar multiplication, is a vector space or not. If not we have to state the axiom that fails.1. u+v is in V. We have, u=x1y1, v=x2y2. So, u+v=x1y1+x2y2=x1+x2y1+y2=xy∈V where x=x1+x2; y=y1+y2. Hence u+v is in V. 2. u+v=v+u We have, u=x1y1, v=x2y2. So,u+v=x1y1+x2y2=x1+x2y1+y2=x2+x1y2+y1 ∵x1+x2=x2+x1& y1+y2=y2+y1=x2y2+x1y1=v+u Hence u+v=v+u.3. (u+v)+w=u+(v+w) We have, u=x1y1, v=x2y2, w=x3y3. Now,u+v+w=x1y1+x2y2+x3y3=x1+x2y1+y2+x3y3=x1+x2+x3y1+y2+y3=x1+x2+x3y1+y2+y3=x1y1+x2+x3y2+y3=x1y1+x2y2+x3y3=u+v+w Thus, (u+v)+w=u+(v+w). 4. There exists an element 0 in V, called a zero vector, such that u+0=u. Now,u+0=x1y1+00=x1+0y1+0=x1y1=u Thus, 0=00 is the zero vector in V.5. For each u in V, there is an element -u in V such that u+(-u)=0 We have,u+-u=x1y1+-x1-y1=x1+-x1y1+-y1=x1-x1y1-y1=00=0 Hence, u+(-u)=0 6. cu is in V. Now,cu=cx1y1=cx1y1 by definition=ky1 cx1=k∈V So, cu is in V.7. c(u+v)=cu+cv Now,cu+v=cx1y1+x2y2=cx1+x2y1+y2=cx1+x2y1+y2 By definition=cx1+cx2y1+y2=cx1y1+cx2y2=cx1y1+cx2y2 By definition=cu+cv Thus, c(u+v)=vu+cv. 8. (c+d)u=cu+du Now,c+du=c+dx1y1=c+dx1y1 By definition=cx1+dx1y1=cx1y1+dx1y1=cx1y1+dx1y1 By definition=cu+du Hence, (c+d)u=cu+du.9. c(du)=(cd)u Now,cdu=cdx1y1=cdx1y1 By definition=cdx1y1 By definition=cdx1y1=cdx1y1 By definition=cdu So, c(du)=(cd)u. 10. 1u=u Now,1u=1x1y1=1x1y1 By definition=x1y1=u Hence 1u=u.Final Answer: All the axioms hold, so the given set is a vector space.