Thread: subspace proof

Let (

V;+;
) be R3 with the usual vector addition and scalar multiplication. For each of
the following, either use the subspace test to show that the given W is a subspace of (V;+;
)
or explain why W is not a subspace.
(i) W := {(x1; x2; x3) eV |x1 + x2 -x3 = 0}.
(ii) W := {(x1; x2; x3) eV |x1 + x2 -x3 = 1}.
(iii) W := {(2t; 3t -1;-t) |t eR}.

(iv) W := {(2t; 3t;-t) |t eR}.

i) W is a subspace, not 100% sure on my working
ii) since 0 vector not included, its not a subspace
iii) Not sure where to start.
iv) Not sure where to start.

I've re-read all my course material over and over but i just need some worked through solutions to get to grips with it. Any help would be much appreciated.
Thanks

i) apply the subspace test. for a,b in R, and (x1,x2,x3), (y1,y2,y3) in W, is a(x1,x2,x3) + b(y1,y2,y3) in W? is W non-empty?

iii) is (0,0,0,0) in W?

iv) apply the subspace test. you can write an element of W as t(2,3,-1), perhaps this will help.

Because a "subspace" has the same operations as the overall space, you don't have to prove things like "existence of a zero", "existence of negatives", etc. still holds. The only things you need to prove are "closure under addition" and "closure under multiplication".

i) If u= (x1, x2, x3) and v= (y1, y2, y3) are in this subspace then x1+ x2+ x3= 1 and y1+ y2+ y3= . Of course, u+ v= (x1+ y1, x2+ y2, x3+ y3). What is (x1+ y1)+ (x2+ y2)+ (x3+ y3)?
If k is a number then ku= (kx1, kx2, kx3). What is kx1+ kx2+ kx3? In order to be in the same subspace, so that we do have closure, those must be 0.

ii) If u= (x1, x2, x3) and v= (y1, y2, y3) are in this subspace then x1+ x2+ x3= 1 and y1+ y2+ y3= 1. Of course, u+ v= (x1+ y1, x2+ y2, x3+ y3). What is (x1+ y1)+ (x2+ y2)+ (x3+ y3)?
If k is a number then ku= (kx1, kx2, kx3). What is kx1+ kx2+ kx3? In order to be in the same subsspace, so that we do have closure, those must be 1.

iii) If u= (x1, x2, y2) and v= (y1, y2, y3) are in this subspace then x1= 2t, x2= 3t- 1, x3= -t, y1= 2s, y2= 3s-1, y3= -s for some numbers t and s. Of course, u+ v= (2t+ 2s, (3t-1)+ (3s-1), (-t)+(-s))= (2(t+ s), 3(+ s)- 2, -(t+s)). Is that of the form (2x, 3x-1, -x) for some number x? Is so, what x? ku= (k(3t), k(2t-1), -kt)= (3(kt), 3(kt)- k, -kt). Is that of the the for (2x, 3x-1, -x) for some number x? If so, what x?

iv) If u= (x1, x2, y2) and v= (y1, y2, y3) are in this subspace then x1= 2t, x2= 3t, x3= -t, y1= 2s, y2= 3s, y3= -s for some numbers t and s. Of course, u+ v= (2t+ 2s, (3t)+ (3s), (-t)+(-s))= (2(t+ s), 3(+ s), -(t+s)). Is that of the form (2x, 3x, -x) for some number x? Is so, what x? ku= (k(2t, 3t, -t)= (2(kt), 3(kt), -(kt))? Is that of the form (2x, 3x, -x) for some number x? If so, what x?

if 0 is not in the subspace, you can't have closure under addtion, because u + (-1)u will not be in the subspace.

you do have to show the subspace actually has SOMETHING in it, as the empty set is not a vector space.

closure of vector addition and closure of scalar multiplication can be combined in a single statement:

for u,v in W, and a,b in F, au + bv is in W. this is probably what his text is referring to as "the subspace test"

(this is analogous to the "subgroup" test for groups H is a subgroup if gh^-1 is in H for all g,h in H. for an abelian group

this becomes g - h is in H. but since a vector space has scalar multiplication, -v is just (-1)v (because of the vector space axiom

(a+b)v = av + bv), so the condition au + bv in W tests for BOTH scalar multiplication and vector addition closures).