Question

Let V = x, yeR. Prove that V is closed for addition and scalar multiplication i.e. that V is the subspace of R.

The original question got deleted for "not having enough information". Ironic how this is the whole question. Mod must've not understood. Here's a repost with my answer.

Answer 1

To prove V is a subspace of R, you must show:

(1) It is not empty

(2) Adding vectors returns real numbers

(3) Multiplying by scalars return real numbers

2 & 3 follow the conditions that x,y must be real.

===

(1):

0 = 0 + 0 , therefor V cannot be an empty set. Really thats it.

===

(2)

Let u=(u1, u2 , 2u1-3u2)

Let v *lowercase =(v1, v2, 2v1-3v2)

Great, now lets add them. This isn't intuitive unless you finish reading it all.

u+v = ( u1 + v1 , u2 + v2 , 2u1 - 3u2 + 2v1 - 3u2 )

--> u + v = [ (u1+v1), (u2+v2) , (2u1 + 2v1) + (-3u2 - 3v2) ] (got like terms kinda)

====> All must equal (x,y, 2x-3y)

... which means x=u1+v1 & y=u2+v2 (you compare u+v to V for this)

===

(3)

Let c be a real number.

C(u) = (Cu1, Cu2, C2u1 - C3u2)

====> Must equal (x,y, 2x-3y)

... which means Cu1 = x & Cu2 = y

Cu is in V (because x,y are)

-> V is a subspace of R3.