Determine whether the given set, together with the specified operations of addit

Question

Determine whether the given set, together with the specified operations of addition and scalar multiplication is a vector space. If it not, select all of the axioms that fail to hold (Let u, v, and w be vectors in the vector space V, and let c and d be scalars.) R^2, with the usual scalar multiplication but addition defined by [X1 Y1] + [X2 Y2] = [X1 + X2 + 1 Y1 + Y2 + 1] All of the axioms hold, so the given set is a vector space. u + v is in V. u + v = v + u (u + v) + w = u + (v + w) There exists an element 0 in V, called a zero vector, such that u + 0 = u For each u in V, there is an element -u in v such that u + (-u) = 0. cu is in V c(u + v) = cu + cv (c + d)u = cu + du c(du) = (cd) u 1u = u

Explanation / Answer

The following axioms hold:

1. u +v is in V

2. u+v = v+u

3. (u+v)+w = u +(v+w)

4. The element 0 = (-1,-1)T is in V . If u = (x1,y1)T, then (u +0) = (x1 -1 +1, y1 -1+1)T = (x1 , y1 )T = u. Thus (-1,-1)T is the 0 vector in V.

5. For a u = (x1,y1)T in V, -u = ( -x1 -2, -y1 -2)T is in V as u +(-u) = (x1,y1)T + (-x1 -2, -y1 -2)T =  (-1 , -1)T i.e. the 0 vector.

6. cu is in V

8. (c+d)u = cu+du

9. cd(u) = c(du)

101u = u .

However, the 7th axionm does not hold. If u = (x1,y1)T and v = u = (x2,y2)T , then cu = (cx1, cy1)T anf cv = (cx2, cy2)T so that cu +cv = ( cx1 + cx2 +1 , cy1+cy2 +1)T . However, c(u+v) = c( x1 +x2 +1, y1 +y2 +1)T =

( cx1+cx2 +c, cy1 +cy2 +c)T. Hence cu +cv c(u+v).

Hence V is not a vector space.

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