Basic Linear Algebra Proof - Orthogonal Vectors Prove that if $\mathbf{u}$ and $\mathbf{v}$ are nonzero orthogonal vectors in $\Bbb R^n$ they are linearly Independent.
I've struggled with this a bit, here is what I know so far:
Suppose $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.  Then $\mathbf{u\cdot v}=0$
and
$c_1\mathbf{u}+c_2\mathbf{v}=0$ is linearly Independent iff $c_1=c_2=0$
I know I need to end with $c_1=c_2=0$ but I can't find a path that reaches this conclusion.  I feel like I need to use properties of the dot product to connect my first assumption to my second assumption but I'm lost on the way.  
 A: 
Then $c_1\mathbf{u}+c_2\mathbf{v}=0$ is linearly Independent iff $c_1=c_2=0$

This does not actually make sense: $c_1\mathbf{u}+c_2\mathbf{v}=0$ isn’t something that can be linearly independent. Linear independence is a property of sets of vectors; $c_1\mathbf{u}+c_2\mathbf{v}=0$ is an equation involving vectors, but it is not a set of vectors.
The correct statement is that the set $\{\mathbf{u},\mathbf{v}\}$ is linearly independent (or, more casually, that the vectors $\mathbf{u}$ and $\mathbf{v}$ are linearly independent) iff it has the following property:

if $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$, then $c_1=c_2=0$.

Your problem, then, is to show that if $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, and if $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$, then $c_1=c_2=0$.
To do this, assume that $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$. Now apply the hint that Ayman Hourieh gave in his answer to show that $c_1=0$ and $c_2=0$.
A: Just use contradiction. If $u,v$ are linearly dependent and non-zero, then $u = cv$ (note: $c$ is a function of your $c_1,c_2$). Then $u\cdot v \neq 0 \rightarrow$ contradiction.
A: Hint: Calculate:
$$
(c_1 \mathbf{u} + c_2 \mathbf{v}) \cdot \mathbf{u}
$$
What do you conclude about $c_1$? Do something similar for $c_2$.
