written 5.1 years ago by | modified 2.2 years ago by |
(1, y) + (1, x) = (1, y+x) and
k(1, y) = (1, ky) is a vector space.
written 5.1 years ago by | modified 2.2 years ago by |
(1, y) + (1, x) = (1, y+x) and
k(1, y) = (1, ky) is a vector space.
written 5.1 years ago by |
Solution:
(a) Closure:
$C_{1}:$ If u=(1, x), v=(1, y) are two element then,
u+v = (1, x)+(1, y) = (1, x+y)
Therefore, v is closed under addition.
$C_{2}:$ If u=(1, x) then,
ku = k(1, x) = (1, kx)
Therefore, v is closed under multiplication.
(b) Addition:
$A_{1}:$ u+v = (1, x)+(1, y) = (1, x+y) = (1, y+x) = (1, y)+(1, x) = v+u
$A_{2}:$ (u+v)+w = [(1, x)+(1, y)]+(1, z)=(1, x+y)+(1, z)
(u+v)+w = (1, (x+y)+z) = (1, x+(y+z)) = (1, x)+[(1, y+z)] = (1, x)+[(1, y)+(1, z)] = u+(v+z)
$A_{3}:$ u+0 = (1, x)+(1, 0) = (1, x+0) = (1, x) = u
0 = (1, 0) is zero for this space.
$A_{4}:$ u+(-u) = (1, x)+(1, -x) = (1, x-x) = (1, 0)
$\therefore$ (1, -x) = -u is additive inverse of (1, x) = u
(c) Scalar Multiplication:
$M_{1}:$ k(u+v) = k[(1, x)+(1, y)] = k(1, x+y) = (1, k(x+y)) = (1, kx+ky) = (1, kx)+(1, ky) = k(1, x)+k(1, y) = ku+kv
$M_{2}:$ (k+l)u = (k+l)(1, x) = (1, (k+l)x) = (1, kx+kl) = (1, kx)+(1, lx) = k(1, x)+l(1, x) = ku+lu
$M_{3}:$ (k.l)u = (k.l) (1, x) = (1, klx) = k(1, lx) = k[l(1, x)] = k(lu)
$M_{4}:$ 1.u = (1, 1.x) = (1, x) = u
Hence, 1 is the multiplicative identity for this space.
Since all the actions are satisfied so V is vector space.