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