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.
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
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).