How to determine the span of two vectors: $(4,2)$ and $(1, 3)$

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.

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, $rm{span}{v_1,dots,v_n}={a_1v_1+dots+a_nv_n|a_iinBbb F, forall i}$.

(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 $Bbb R^2$.)

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 $a,binmathbb{R}$.

So for example $(5,5)$ is in the span of your vectors, because $1cdot (4,2)+1cdot (1,3)=(5,5)$

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.