$V$ is a vector space with inner product $( , )$. If $(v,w)=0$ for all $v$ belongs to $V$, show that $w=0$.

Question

I found a solution that let $v =(x_1, x_2,...,x_n)$ and $w=(y_1, y_2,...,y_n)$. Then $(v,w)= x_1*y_1+...+x_n*y_n = 0$. The only way that this is equal to $0$ is when $y_1=y_2=...=y_n=0$. Hence $w=0$. But I don't think that this is a solid proof. Is there any other way to prove this like using any identities etc?

Answer 1

Here is a solution which does not depend on coordinates or components of vectors:

The axioms defining a standard inner product $(\cdot, \cdot)$ include

$\forall w \in V, \; (w, w) \ge 0, \tag 1$

and

$(w, w) = 0 \Longleftrightarrow w = 0; \tag 2$

if

$\forall v \in V, \; (v,w) = 0, \tag 3$

then taking

$v = w, \tag 4$

we have

$(w, w) = 0, \tag 5$

whence by (2),

$w = 0. \tag 6$