3.a) suppose c1(v1+v2) + c2(v2+v3) + c3(v1+v3) = 0

collecting terms, we have:

(c1+c3)v1 + (c1+c2)v2 + (c2+c3)v3 = 0

and so by the linear independence of {v1,v2,v3}:

c1+c3 = 0

c1+c2 = 0

c2+c3 = 0

hence c3 = c2 = -c1 (equations 1&2), so c2+c3 = -2c1 = 0

now, IF the characteristic of F is NOT 2, then -2 ≠ 0, and we can divide by it (this is where we use the fact that char(F) ≠ 2).

this gives us c1 = 0, and c2 = c3 follows immediately, which proves the linear independence of T.

3.b) if you read the above carefully, you will see that if char(F) = 2, we can find non-zero elements c1,c2,c3 with c1(v1+v2) + c2(v2+v3) + c3(v1+v3) = 0

(try some "simple" values for c1,c2,c3).

4. this is very simple. if c1v1 + c2v2 + ...+ cnvn = 0 implies ALL the cj are 0,

then suppose that our subset is the first k of the vj (by re-ordering them if needed).

then we have c1v1 + c2v2 +...ckvk = 0, so

c1v1 + c2v2 +...+ ckvk + 0v(k+1) +...+ 0vn = 0, which by the linear independence of the entire set means....?

5. suppose we have c1v1 + c2v2 +...+ cnvn + dv = 0

there are two cases: d = 0, and d ≠ 0.

suppose d ≠ 0. then -dv = c1v1 + c2v2 +...+ cnvn, so

v = (-c1/d)v1 + (-c2/d)v2 +...+ (-cn/d)vn, which means v is in span(S), but....(finish this)

but if d = 0, what can we say about the remaining sum?