Linear Algebra, proving subset is a subspace 
Let $W$ be a subset of vector space $V$ over $K$. $\forall \mathbf{u}, \mathbf{v} \in W,\alpha \in K, \alpha \mathbf{u} + \mathbf{v} \in W$ , show that $W$ is a subspace over $K$. Hence, show that the set of linear combinations
  $$W = \{\alpha_1 \mathbf{v}_1 + \alpha_2\mathbf{v}_2 +\ldots + \alpha_n\mathbf{v}_n, \mathbf{v}_i \in V, \alpha_i \in K,i = 1, \ldots, n\}$$
  is a subspace of $V$ over $K$.

I don't understand what $W$ contains? Is $W = \{u,v\}$? or is $W = \operatorname{Span}\{u,v\}$??
Please help, I am new to linear algebra.
 A: You are not given what $W$ contains. And that is the whole point; you don't need to know what exactly $W$ contains, as long as it contains anything, and the given condition holds, you already can tell that it is a subspace, even though you don't know what it is.
Indeed, this is exactly where the power of such theorems comes from: They allow you to make conclusions even if you know very little about the set in question.
Also note that $\mathbf u$ and $\mathbf v$ are quantified, "for all $\mathbf u$ and $\mathbf v$ in $W$". That is, the statement is not about two specific, given vectors, but it means that if you draw any two vectors from $W$, the condition must hold (including, but not limited to, the case that you select the very same vector twice).
So your task is to prove that every non-empty subset $W$ that fulfils the given condition is a subspace of $V$.
(Note that I added “non-empty” to the claim because otherwise the statement you are supposed to prove is not true, as the empty set is not a vector space; if the task was actually given as stated, you might want to give that counterexample, and then proceed to prove the statement for non-empty sets).
A: To check that $W$ is a vector subspace you need to check the 3 following conditions: i) $W$ is non empty (clear if $V$ is non empty), ii)if $\mathbf x \in W$ and $\mathbf y \in W$, then $\mathbf x+\mathbf y \in W$. iii)If $\alpha \in K$, and $\mathbf x \in W$, then $\alpha \mathbf x \in W$
For your second question, you need to check these three conditions again. Again (i) should hold.
For ii) and iii), observe that: for $x_i$ and $y_i$ in $K$, and $v_i$ in $V$, we have:
$(x_1 v_1+...+x_n v_n)+(y_1 v_1+...+y_n v_n)=(x_1+y_1)v_1+...+(x_n+y_n)v_n$
, which is again a linear combination of the $v_i$'s.
Now check that: for $c$ in $K$:
$c(x_1 v_1+...+x_n v_n)=cx_1 v_1+...+cx_n v_n$ is again a linear combination of the $v_i$s.
A: Without more information about a basis for W, we can't recognise what does W contain.
But if we have more information about W and V, for example, let a basis for W to be $\{\underline{u},\underline{v}\}$ and a basis for V to be 
$\{\underline{u},\underline{v},\underline{w},\underline{x}\}$

