The easiest way to answer this question is show that a linear combination of the vectors yields the generating vectors of S, i.e.uandw. In that case, say and for some scalars a,b,c,d and vectors x,y. Then , 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 . Since q=-2w, , so x is in S.