Linear Algebra ! Proofs of Span Problems

Question

Please don't give me the detailed answer but I am looking for some guidance. Okay the question is in this link (https://i.stack.imgur.com/FVqCP.jpg).

The question is : 1a. Prove that Span (v) is a subset of Span (u,v) and Span (v,w) for every u,v,w in R^n

1b.Let u,v,w be elements in R^n. Assuming u is not an element of span(w) and v is not element of span(u,w). Show that span(v) = span(u,v) and span (v,w).

Part a) is really simple and I already have a proof for that.

Its part b) where I just want CONFIRMATION that I am on the right path. So the proving the subset relation in the left-right direction is just part A.

To prove the other direction this is what I have done so far(https://i.stack.imgur.com/g3khy.jpg). I am now trying use the given assumptions.

Assumption 1. $u$ is not an element of span $w$ and

2. $v$ is not an element of $u$ and $w$.

Okay using 1, I can conclude that $ew - bu$ cannot be equal to zero since you can express u as a linear combination of $w$. Furthermore since you can't express v as a linear combination of $u$ and $w$ then $c-d$ has to equal zero. Thats as far as I can go. My question is am I on the right track and if so could you in the vaguest possible terms till me where to go from there? Thank for the assistance and also the kinda stringent conditions but yeah...

Answer 1

If $x$ is an element of the right-hand side, then you can write $$x = c_1 u + c_2 v = c_3 v + c_4 w$$ for some scalars $c_1,\ldots, c_4$. Try to use the conditions in the problem to deduce that $c_1 = c_4 = 0$.

We have $(c_2 - c_3)v = -c_1 u + c_4 w$ which implies $(c_2 - c_3)v \in \text{span}\{u, w\}$. If $c_2 \ne c_3$, this contradicts $v \notin \text{span}\{u,w\}$, so we must have $c_2=c_3$. Then we have $c_1 u = c_4 w$, and the other condition yields...