How to show the if a list of vector spans $V$ Show that the list $(1,2), (3,5)$ is a basis for $\mathbb R^2$.
In order to show it is a basis, I have show that the list is linearly independent and that the list spans $\mathbb R^2$. 
So I understand how to show linear independence (you simply set up a system of equations). However, I am uncertain as to how to show that it spans $\mathbb R^2$. How can you be certain that just by using $(1,2)$ and $(3,5)$ you will be able to produce all $\mathbb R^2$?
 A: To show that $(1,2),\ (3,5)$ span $\mathbb R^2$, you need to show that given $(a,b)\in \mathbb R^2$, we can find $\alpha,\beta\in\mathbb R$ such that $$(a,b) =\alpha(1,2)+\beta(3,5)$$
or equivalently, we need to solve the simultaneous equations
$$a = \alpha + 3\beta\\ b=2\alpha + 5\beta.$$
One can manually show that these simultaneous equations always have solutions for any $(a,b)$, either by solving as usual, or by using the fact that the matrix
$$\begin{pmatrix}1&3\\2&5\end{pmatrix}$$is invertible.
A: The determinant of $$\begin{pmatrix}1&3\\2&5\end{pmatrix}$$
is nonzero over any field $F$. Hence the vectors are a basis for $F^2$. In particular this is true for $F=\mathbb{R}$.
A: A linearly independent list with the same cardinality as a basis is a spanning set, so a basis. (Assuming finite dimensional spaces, of course.)
This is because every linearly independent list can be extended to a basis; so if yours wasn't a basis, you'd end up with a basis of $R^2$ with more than two elements.
A: There's an easier way to show that it is a basis of $R^2$. 
There's a sentence that states that if you can show a a group of vectors are linearly independent,  and the dimension of the group is equal to the dimension of the space, then that group is a basis for the space.
