Is the set of all vectors in the span of $\pmatrix{1 \\ 2}$ and $\pmatrix{2 \\ -1}$ a subspace or not? I'm trying to figure out whether or not the set of all vectors in the span of $\pmatrix{1 \\ 2}$ and  $\pmatrix{2 \\ -1}$ is a subspace or not (I know the answer to be yes, it is a subspace but I want to understand how to prove it).
So far I have:
$U=(1(X_1),2(Y_1))$
$V=(2(X_2),-1(Y_2))$ 
$U+V=(1(X_1) , 2(Y_1)) + (2(X_2) , -1(Y_2))=(3(X_1+X_2) , 1(Y_1+Y_2))$
Not sure if I'm going down the right path here $\dots$
Thanks in advance
 A: Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W=  \left\{ a \pmatrix{1 \\2}+b\pmatrix{2 \\-1} \big| a, b \in \mathbb{R} \right\} $$
Subspaces must satisfy the subspace test, that is a subspace


*

*Is nonempty (Often we show it contains the zero vector, this is a nice test).


Clearly if we take $a=b=0$ the zero vector is contained in this set.


*Is closed under linear combinations.


You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 \pmatrix{1 \\2}+b_1\pmatrix{2 \\-1}$ and $w=a_2 \pmatrix{1 \\2}+b_2\pmatrix{2 \\-1}$ in $W$, we now consider:
$$ \lambda v + \mu w= \lambda (a_1 \pmatrix{1 \\2}+b_1\pmatrix{2 \\-1})  + \mu ( a_2 \pmatrix{1 \\2}+b_2\pmatrix{2 \\-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ \lambda v + \mu w= (\lambda a_1 + \mu b_1) \pmatrix{1 \\2}+(\lambda a_2 + \mu b_2)\pmatrix{2 \\-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
