Then prove: $\vec{v} = \vec{0}$ if $\langle u,v\rangle = 0$ If $\vec{v} \in V$ such that $\langle u,v\rangle = 0$, $\forall \vec{u} \in V$.
Then prove: $\vec{v} = \vec{0}$
I tired to solve by assuming that they are $\langle u,v\rangle \neq 0$ $\rightarrow$ $u,v$ are not orthogonal, and if they are not orthogonal then $\langle u,v\rangle = 0$ can only be true if and only if $\vec{v} = \vec{0}$.
Is this correct? Or is there any other alternative way?
 A: Easiest way:  pick $u = v$: then
$\Vert v \Vert^2 = \langle v, v \rangle = 0,  \tag{1}$
which implies
$\Vert v \Vert = 0, \tag{2}$
i.e.
$v = 0. \tag{3}$
QED!
Note Added Thursday 16 April 2015 3:23 PM PST:  The usual definition of an inner product on a vector space $V$ specifies that $\langle v, v \rangle \ge 0$, $\forall v \in V$ with $\langle v, v \rangle = 0$ iff $v = 0$.  This allows the definition of $\Vert v \Vert = \langle v, v \rangle^{1/2}$ to proceed, with the consequence that $\Vert v \Vert = 0$ precisely when $v = 0$. These notions are reasonably widely known, and assumed so here.  See  http://en.m.wikipedia.org/wiki/Inner_product_space.  End of Note.
A: $$
u = v\in V \Rightarrow 0 = u \cdot v = v\cdot v = \lVert v \rVert^2 \Rightarrow \lVert v \rVert = 0 \Rightarrow v = 0
$$


*

*property of task

*definition scalar product $a \cdot b = \lVert a \rVert \, \lVert b \rVert \cos \angle(a,b)$

*property of squares

*property of vector norm
A: If  v is a fixed vector and $\langle u,v\rangle=0$ for all u $\in V$, then in particular for $u=e_i$ for $i=1,\ldots,n$.
We know also that $v=a_1e_1+\ldots+a_ne_n$
 Then $\langle e_1,v\rangle=0=a_1\langle e_1,e_1\rangle+\ldots+a_n\langle e_1,e_n\rangle=a1\langle e_1,e_1\rangle$ 
but $\langle e_1,e_1\rangle \neq 0$ so $a_1=0$. 
Make an analogous process with all $e_i$ and you will find that $v=0$.
