Span of two vectors with positive coefficients - Linear system with positive solution Given any two linearly independent vectors $\vec v_1,\vec v_2\in \mathbb{R}^2$, I'm interested in their linear span with positive coefficients. Given any other vector $\vec v\in \mathbb{R}^2$, how can one easily see if there exist $\alpha_1,\alpha_2\in \mathbb{R}^+$ such that $\vec v=\alpha_1\vec v_1+\alpha_2\vec v_2$ without solving the linear system?
This is equivalent to ask which is a necessary and sufficient condition for the linear system
$$x=\alpha_1x_1+\alpha_2x_2\\
y=\alpha_1y_1+\alpha_2y_2$$
to admit positive solutions $\alpha_1,\alpha_2$
 A: For a clear explanation, we may start to look at what exactly the coordinate is.
You know that when it comes to span $\mathbb{R}^2$ with a set of two linearly independent vectors $\mathbf{v},\mathbf{w}$, for every vectors in $\mathbb{R}^2$, you could find two real number $x ,y$ such that it could be expressed as $x \mathbf v +y\mathbf w $. And now I claimed: as under the standard orthogonal basis $\{ (0,1)^T , (1,0)^T \}$, the corresponding $(x,y)$ is what we used as coordinates, and now if generalize it, we say the set$(x,y)$ under the basis $B=(\mathbf v ,\mathbf w )$ is called the coordinate system under $B$.
If you get this, then the problem is going to be trivial. What you need is just $(x,y)$ under $B$ to be positive, which is equivalence to all the vectors in the Quadrant I of $B$(recalls similar definition as under standard orthogonal basis), that is, in a easy way, when the plane $\mathbb{R}^2$ is cut into four part by the spanning sets of $\mathbf v$ and $\mathbf w$ respectively, Quadrant I is the part that $\mathbf v +\mathbf w $ lies on.
