# Suppose that v and w are vectors in a vector space V . Suppose that S is a subspace of V . Let T be the subspace spanned by v and S. Let U be the subspace spanned by w and S. Prove that if w is in T but not in S, then v is in U

I'm not sure that equation (1) has to be true. Let `V=RR^2, S=span{(1,0)}, v=(0,1), w=(1,1).` Then all conditions hold but `w` is not a multiple of `v.`

What is true is that if `{s_1,s_2,...,s_n}` is a basis for `S` (so we'll assume `S` is finite dimensional), then

`w=alphav+alpha_1s_1+...+alpha_ns_n,`  where `alpha!=0,` since otherwise...

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I'm not sure that equation (1) has to be true. Let `V=RR^2, S=span{(1,0)}, v=(0,1), w=(1,1).` Then all conditions hold but `w` is not a multiple of `v.`

What is true is that if `{s_1,s_2,...,s_n}` is a basis for `S` (so we'll assume `S` is finite dimensional), then

`w=alphav+alpha_1s_1+...+alpha_ns_n,`  where `alpha!=0,` since otherwise `w` would be in `S.` In that case, we have

`v=1/alpha(w-a_1s_1-a_2s_2- ... -a_ns_n),` which is in `U` by definition.

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Since `T` is spanned by `v` and `S` and `w` is in `T` but not in `S` it follows that

`w=alpha v`                                                                           (1)

for some real number `alpha.` Then we also have

`v=1/alpha w`                                                                             (2)

So because of (1) and (2) `v` is in any space spanned by `w` and vice versa. And since `U` is spanned by `w` and `S` because of (2) `w in U`.

Furthermore you can see that `T=U` because they are both spanned by `S` and `v` and `w` are (because of (1)) from the same subspace of `V.`

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