Proof involving the span of a set of vectors. Hi I'm stuck on a proof homework question. The question is:
"Suppose that $u,w_{1},...,w_{l} ∈ \mathbb R^{n}$. Prove that span{$w_{1},...,w_{l},u$} = span{$w_{1},...,w_{l}$} if and only if $u$ is a linear combination of $w_{1},...,w_{l}$."
Below is what I've done so far. Is it valid? I feel it isn't as it was too simple.
(https://i.stack.imgur.com/iT5Xs.jpg)
 A: '$\Rightarrow$': $\text{span}\{w_1, w_2, ..., w_l,u\}$ = $\text{span}\{w_1, w_2, ..., w_l\}$ means that any vector that can be written as a linear combination of one of the two sets can also be written as a linear combination of the other. Choose $v \in$ $\text{span}\{w_1, w_2, ..., w_l,u\}$ with $v = u+\sum_{i=1}^{l}\alpha_iw_i$. $v$ can also be written as $v=\sum_{i=1}^l\beta_iw_i$, which means $u=\sum_{i=1}^{l}(\beta_i-\alpha_i)w_i$.
'$\Leftarrow$': $u$ is a linear combination of $\{w_1, w_2, ..., w_l\}$ means it can be written as $u=\sum_{i=1}^{l}\lambda_iw_i$.


*

*Choose an arbitrary $v \in$ $\text{span}\{w_1, w_2, ..., w_l\}$ and write it
as $v=\alpha_uu+\sum_{i=1}^{l}\alpha_iw_i =
   \sum_{i=1}^{l}(\alpha_u\lambda_i+\alpha_i)w_i$. Hence:
$\text{span}\{w_1, w_2, ..., w_l,u\}$ $\subseteq$ $\text{span}\{w_1, w_2, ..., w_l\}$.

*Choose an arbitrary $v \in$ $\text{span}\{w_1, w_2, ..., w_l\}$ and write it
as $v=\sum_{i=1}^{l}\alpha_iw_i$. This is a valid linear combination of $\{w_1, w_2, ..., w_l, u\}$, which means
$\text{span}\{w_1, w_2, ..., w_l\}$ $\subseteq$ $\text{span}\{w_1, w_2, ..., w_l,u\}$.
It follows that the two sets span the same vector space.
A: First $(\implies)$
Let's take $x \in$ span$\{w_1,...w_l,u\}=$ span$\{w_1,...w_l\}$, so:
$$x=a_1w_1+a_2w_2+...+a_lw_l+pu=b_1w_1+b_2w_2+...+b_lw_l$$
Such that $p \ne 0$, we conclude:
$$u=\left(\frac{b_1-a_1}{p}\right)w_1+...+\left(\frac{b_l-a_l}{p}\right)w_1$$
Second $(\impliedby)$
Let's take $x \in$ span$\{w_1,...w_l,u\} \Rightarrow x=a_1w_1+a_2w_2+...+a_lw_l+pu$, but $u=b_1w_1+b_2w_2+...+b_lw_l$, then:
$$x=(a_1+p.b_1)w_1+(a_2+p.b_2)w_2+...+(a_l+p.b_l)w_l$$
what give us that $x \in$ span$\{w_1,...w_l\}$.
Reciprocally, let's take $x \in$ span$\{w_1,...w_l\} \Rightarrow x=a_1w_1+a_2w_2+...+a_lw_l$, but $u=b_1w_1+b_2w_2+...+b_lw_l$, then:
$$x=(a_1-b_1)w_1+(a_2-b_2)w_2+...+(a_l-b_l)w_l+u$$
what give us that $x \in$ span$\{w_1,...w_l,u\}$.
A: Here is how I would prove it.  
If $\text{span}\{w_{1},\ldots,w_{k},u\}=\text{span}\{w_{1},\ldots,w_{k}\}$, then since $u$ (obviously) belongs to the LHS, it must belong to the RHS; that is $u \in \text{span}\{w_{1},\ldots,w_{k}\}$, so by definition $u$ is a linear combination of $w_{1},\ldots,w_{k}$.  
Conversely, if $u$ is a linear combination of $w_{1},\ldots,w_{k}$, say $u=\sum_{i=1}^{k}a_{i}w_{i}$, then any linear combination of $w_{1},\ldots,w_{k},u$ can be written
$$\sum_{i=1}^{k}b_{i}w_{i}+cu=\sum_{i=1}^{k}(b_{i}+ca_{i})w_{i}$$
So $\text{span}\{w_{1},\ldots,w_{k},u\}\subset \text{span}\{w_{1},\ldots,w_{k}\}$. But it's clear that $\text{span}\{w_{1},\ldots,w_{k}\}\subset \text{span}\{w_{1},\ldots,w_{k},u\}$ too, so they must be equal.
