Proving a set is a basis, having already given a basis Question: If {$u,v$} is a basis for the subspace U, show that {$u+2v,-3v$} is also a basis for U
My attempt: 
We must prove that {$u+2v,-3v$} spans U and is linearly independent
We know that given any $w \in U$ there exists $w = a_1(u) + a_2(v)$ where $a_1,a_2 \in \Bbb F$ and we also know that if $a_1(u) + a_2(v) = 0$ then $a_1=a_2=0$
Therefore: given any $w \in U$ there exists $w = a_1(u+2v) + a_2(-3v)$, which can be written as $w = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$ therefore {$u+2v,-3v$} spans U
By the same principle, $0 = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$, 
let $a_3= 2a_1-3a_2 $, we can rewrite as $0 = (a_1)u + a_3(v)$
Therefore {$u+2v,-3v$} is also linearly independent, therefore a basis for U.
Is this correct? Thank you very much! 

There are several issues with your presentation and proof. You're talking about a subspace without mentioning the ambient space. Also, it isn't true over arbitrarily fields, e.g. it fails over a field of characteristic $3$.
I'll assume you're working over a field of characteristic $\neq 3$ and that you want to prove that $\{u,v\}$ being a basis for a vector space implies that $\{u + 2v,-3v\}$ is also a basis. Since you already know that your space is $2$-dimensional, it suffices to show that $\{u + 2v,-3v\}$ is linearly independent. So suppose $0 = a(u+2v) -3bv = au + (2a-3b)v$. Linear independence of $u,v$ implies that $a = 0$ and $2a - 3b = 0$, so $b = 0$ also, which is what we wanted to show.
A more general way to automate these kinds of arguments is to map your basis to your new suggested basis and prove that the map is an isomorphism, using your preferred method (Gauss-Jordan reduction, determinant $\neq 0$, ...).
