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! 
 A: 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 $\lambda_1=\lambda_2=0$. 
A: 
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

You should not write "given $w\in U$, there exists $w = \dots$", because $w$ exists, you are given it. I think you mean "given $w\in U$, there exists $a_1$, $a_2$ so that $w = \dots$". 
Also, you are trying to show that there exist $a_1$, $a_2$ so that $w = a_1 (u+2v) + a_2(-3v)$. You do not know it yet. What you do know is that $w = b_1 u + b_2 v$. You need to play around with this so that you get $w = a_1 (u+2v) + a_2(-3v)$.

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 $0 = (a_1)u + a_3(v)$

This one is better, but still needs work. If $0 = a_1 (u +2v) + a_2 (-3v)$, then $0 = a_1 u + (2a_1-3a_2) v$. At this point, we know that since $u$ and $v$ are linearly independent, we can say $a_1 = 0$ and $a_3 := 2a_1 - 3a_2 = 0$. Remember that you wanted to show that $a_1 = a_2 = 0$ and you've only showed the first part so far. The next step is to say that because $a_3 = 0$ and $a_1=0$, we can write $a_3 = 2a_1-3a_2$ as $0 = -3a_2$, so $a_2 = 0$. Now, you know that they are linearly independent.
Hopefully seeing the work written out in the second part will give you an idea of how to work on the first part.
A: 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 $\neq 0$, ...).
A: To prove that $u+2v$ and $-3v$ are linearly independent we take a linear combination equated to zero, that is
$$A(u+2v)+B(-3v)=0,$$
and deduce $A=B=0$. 
But $A(u+2v)+B(-3v)=Au+(2A-3B)v$ then from $Au+(2A-3B)v=0$ we infer that for the coefficients $A=0$ and $2A-3B=0$ since $u,v$ are linearly independent, so $A=0$ and $B=0$. 
A: Use the fact that :Let  $V$ is a vector space 

If $\dim V=n$ then any set of $n$ vectors which are linearly independent is a basis of $V$.

So in your case it is enough to check linear independence only.
