# Thread: Confused with Simple Vector Spaces

1. ## Confused with Simple Vector Spaces

Hi,

I have the vectors:

u = [1 2 1] w = [-1 0 1] q = [2 0 -2] r = [0 2 2]

and I have the vector space:

S = { au + bw | a, b R }

I know how to answer questions which ask if a vector belongs to S - e.g. (q = -2w)

But I do not know how to answer these questions:

(b) Do w and r span the same subspace S?
(c) Do w and q span S?
(d) Do u, w and q span S?

I have no idea how to figure the above out being as my my material is vague and I have tried to search the internet for answers but don't understand any of it.

2. The easiest way to answer this question is show that a linear combination of the vectors yields the generating vectors of S, i.e. u and w. In that case, say $u=ax+by$ and $w=cx+dy$ for some scalars a,b,c,d and vectors x,y. Then $mu+nw=max+mby+ncx+ndy=(ma+nc)x + (mb+nd)y$, or every vector in the span of u and w lies in the span of x and y. Further, if we cannot find a linear combination for the generators, then we have shown the spans don't coincide.

then for (b), the question is asking if span{w,r} = span{u,w}. Since w is a generator of both sets, it remains to show that u is a linear combination of r and w, and r is a linear combination of u and w. It is easy to see that r=u+w,
so u=r-w thus we have acquired both of the desired linear combinations and by the previous remarks the spans form the same subspace.

For (c), can w and q generate u? Note that the second coordinate is 0 for w,q, and any linear combination of these two vectors, but not for u.

Finally for (d), this question amounts to figuring out if there are any vectors outside of S that can be expressed as a linear combination of u,w, and q. But since q is just a multiple of w, clearly we can't get any new vectors! Algebraically, suppose $x=au+bw+cq$. Since q=-2w, $x=au+bw-2cw=au+(b-2c)w$, so x is in S.