TITLE: "Vanishing inner product implies orthogonality" Is it a definition or theorem?
QUESTION [6 upvotes]: The definition of inner product is $\langle u,v\rangle=u_1\bar{v_1}+\cdots+u_n\bar{v_n}$. Two vectors $u,v \in V$ are said to be orthogonal if $\langle u,v\rangle=0$, where $V$ is complex vector space. But why is it true? A lot of books just present this fact without giving any proof. Why such an expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}=0$ will lead to the fact that $u$ and $v$ are orthogonal? Is it an intuition behind the expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}$ that can explain orthogonality?
I read some proof about it but they don't seem correct. For example, some said that $\langle u,v\rangle=|u||v|\cos\theta$ and thus $\theta=90^\circ$. But why are they equivalent?
Some use the Pythagoras theorem, $$\begin{array}{rl}
|u+v|^2 &=\langle u+v,u+v\rangle\\
&=|u|^2+2\langle u,v\rangle+|v|^2
\end{array}$$
So dot product $= 0$ implies orthogonality, but this only works for real vector space. So why exactly inner product $= 0$ implies orthogonality?

REPLY [2 votes]: After some research I found another way to prove this, which in my opinion is the most elementary one: 
Suppose you have two vectors $v,u\in\mathbb{R}^n$, and you want to determine their "angle". Supposing that their angle is 90 degrees, we could apply Pythagoras in order to calculate the area of $|u-v|^2$. This yields, 

$||v||^2+||u||^2=||u-v||^2$, 

which is equivalent to 

$v_1u_1+...+v_nu_n=0$. 

In the two dimensional case we could have argued that $u$ and $v$ has the slopes $t_1$ and $t_2$ respectively, and by applying the same argument you arrive to the condition that $t_1t_2=-1$ (this follows if you take $u=(1,t_1)$ and $v=(1,t_2)$). The last case was already mentioned by AOritz. Actually, I am still trying to figure out how the "slope" argument could be extended to three dimensions but I cannot see this.