Then prove: $\vec{v} = \vec{0}$ if $\langle u,v\rangle = 0$

Question

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?

Answer 1

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.

Answer 2

$$ u = v\in V \Rightarrow 0 = u \cdot v = v\cdot v = \lVert v \rVert^2 \Rightarrow \lVert v \rVert = 0 \Rightarrow v = 0 $$

1. property of task

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

3. property of squares

4. property of vector norm

Answer 3

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 $v=0$.