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! 

First, note that $\dim U=2$ so it suffices to show that $\{u+2\,v,-3\,v\}$ is linearly independent. To do so, note that 
$$
\lambda_1(u+2\,v)+\lambda_2(-3\,v)=0
$$
if and only if
$$
\lambda_1u+(2\,\lambda_1-3\,\lambda_2)v=0\tag{1}
$$
But $\{u,v\}$ is linearly independent so (1) holds if and only if
\begin{array}{rcrcrcrc}
0 &=&\lambda_1\\
0&=&2\,\lambda_1&-&3\,\lambda_2
\end{array}
which holds if and only if $\lambda_1=\lambda_2=0$. 
