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! 


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.
