Linearly Independent (Vectors) If $u$ and $v$ are linearly independent vectors, then does that mean $u+v$ is also linearly independent?
I'm thinking yes, because if $u$ and $v$ are linearly independent, they span all of $\mathbb{R}^2$. So the sum of them most likely would as well. So true. How would you prove it?
 A: Assuming I am reading the question correctly, you want to show that if $u$ and $v$ are independent, then so is $u+v$. So you want to show that (by the definition of linear independence) that $x_1u+x_2v=0\implies x_1=x_2=0$.
Does $c_1(u+v)=0\implies c_1=0?$ Of course! This is because any set that has exactly one non-zero vector is independent. You can think of it this way: Let $u+v=w$, where $w$ is non-zero. Then you are just asking if $c_1w=0\implies c_1=0$. Since $w$ is non-zero, then $c_1=0$. 
A: Your reasoning is off:
If $u$ and $v$ are vectors, linearly independent or not, $u+v$ is just going to be another vector.  It's impossible to span all of $\mathbb{R}^2$ with $u+v$, a single vector, since $\mathbb{R}^2$ is two dimensional.
If you can show that $u+v$ is non-zero though, then you'll have established that the set $\{u+v\}$ is "linearly independent", which can be said for any non-zero vector.  It is not clear whether that is what you're trying to show.
A: Both $\{u, u+v\}$ and $\{v, u+v\}$ are independent sets given the independence of $\{u, v\}$ however the set containing all three $\{u, v, u+v\}$ is a linearly dependent set.
Proof of one of them, Assume that whenever $a_{1}u + a_{2}v = 0$ then it necessarily follows that $a_{1} = a_{2} = 0$ and that $c_{1}u + c_{2}(u + v) = 0$ is given. Then we can rewrite it as $(c_{1} + c_{2})u + c_{2}v = 0$ implying that $c_{1} + c_{2} = c_{2} = 0$. And with $c_{2} = 0$ we additionally conclude $c_{1} = 0$ completing the proof of the independence of $\{u, u+v\}$.
The remaining claims follow similarly. 
