How to determine the span of two vectors: $(4,2)$ and $(1, 3)$ How to determine the span of two vectors in $\mathbb R^2$:
$(4,2)$ and $(1, 3)$
Do I subtract them? I don't how I'd solve this. Thanks in advance. In my question the vectors are like this:
\begin{bmatrix}
4\\
2
\end{bmatrix}
But that doesn't matter, right?
Would the vector equation $x_1v_1 + x_2v_2$ = b be consistent for any b in $\mathbb R^2$
?
 A: The span is just the possible linear combinations of the two vectors...
$$Span\{(4,2),(1,3)\}=\{a(4,2)+b(1,3);a,b\in \mathbb{R}\}$$
A: The span of a set of vectors, is the set of every linear combination that you can "create" from those vectors.
So in your example $a(4,2)+b(1,3)$, where $a,b\in\mathbb{R}$.
So for example $(5,5)$ is in the span of your vectors, because $1\cdot (4,2)+1\cdot (1,3)=(5,5)$
Also $(3,-1)$ is in the span as $(4,2)-(1,3)=(3,-1)$.
In general every vector of the form $(4a+b,2a+3b)$ are in the span.
A: Hint:  You can write the vectors vertically or horizontally, it really doesn't matter.
The span of any $n$ vectors $v_1,\dots,v_n$ is by definition the set of all linear combinations of them, $\rm{span}\{v_1,\dots,v_n\}=\{a_1v_1+\dots+a_nv_n|a_i\in\Bbb F\, \forall i\}$.
(This generalizes to infinite dimensions as well.)
(Also, if the vectors are linearly independent, you get a copy of $\Bbb F^n$.  So in this case you get the whole $\Bbb R^2$.)
A: As others have said (or suggested), since (4,2) and (1,3) are linearly independent, their span equals all of $R^2.$
By linear independence, I mean that (for example) there is no scalar 
$k \in \mathbb{R} \;\ni \;(4,2) = k \times (1,3).$
The following analysis proves that any element (x,y) that is in $R^2$ is also 
in the span of (4,2) : (1,3).
Desired to find scalars $r,s$ such that 
$[E_1] \;r \times (4,2) \;+\; s \times (1,3) \;=\; (x,y).$
By $E_1,$ 
$[E_2] \;r(4) + s(1) = x$ and 
$[E_3] \;r(2) + s(3) = y$.
Multiplying $E_3$ by 2 and subtracting it from $E_2$ gives 
$s(-5) = (x - 2y)$. 
Similarly, multiplying $E_2$ by 3 and subtracting it from $E_3$ gives 
$r(-10) = y - 3x.$
By the above analysis, it has been demonstrated that for any $(x,y)$ in $\mathbb{R^2},$ 
there exist scalars $r = (-1/10)(y - 3x)$ and 
$s = (-1/5)(x - 2y)$ 
such that $E_1$ is satisfied.
