Relation between linear independence and inner product I was given the following question:

Let $V$ be an inner product space and let $u,v\in V$ be two nonzero vectors. Prove or disprove: 
  
  
*
  
*If $\langle u,v\rangle=0$, then $u,v$ are linearly independent.
  
*If $u,v$ are independent, then $\langle u,v\rangle=0$. 
  



*

*I know that $u,v$ are arthogonal if $\langle u,v\rangle = 0$.
So, since $\langle u,v\rangle = 0$, and $u,v$ are non zero vectors can I claim linear independence between the vectors directly? And if so, how do I explain it?

*This just seems wrong... I don't see how linear independence leads to this vectors having inner product of zero, meaning they are orthogonal.
Any help or direction would be very helpful.
 A: *

*It is true that $\textbf{u}$ and $\textbf{v}$ are linearly independent with these assumptions, however, it is not sufficient to claim it based on intuition. You can show it as follows:


Suppose $\alpha\textbf{u} + \beta\textbf{v} = \textbf{0}$. Then 
$$
 0 = \langle \textbf{v}, \textbf{0}\rangle = \langle \textbf{v}, \alpha\textbf{u} + \beta\textbf{v} \rangle = \alpha \langle \textbf{v}, \textbf{u}\rangle + \beta\langle\textbf{v}, \textbf{v}\rangle = 0 + \beta|\textbf{v}|^2.
$$
You can conclude from here that $\beta = 0$ (why?). A similar calculation shows that $\alpha = 0$, from which you can conclude that $\textbf{u}$ and $\textbf{v}$ are linearly independent


*For part two it should be easy to come up with a counter example in $\mathbb{R}^2$ to find two linearly independent vectors that are not orthogonal. This will show the statement is false. 

A: No, you cannot claim linear independent directly. You have to prove it. if $\alpha u +\beta v=0$, then 
$$
0=\langle \alpha u+\beta v,u\rangle=\alpha\,\langle u,u\rangle.
$$
As $\langle u,u\rangle\ne0$, it follows that $\alpha=0$. A similar argument shows that $\beta=0$, and so $u,v$ are linearly independent. 
The second statement is wrong indeed. Let $V=\mathbb R^2$ with the usual inner product, let $u=(1,0)$ and $v=(1,1)$. Then $u,v$ are linearly independent but $\langle u,v\rangle=1$. 
