Can $\langle x,x \rangle$ be less than zero on an inner product space? We define an inner product on a vector space $V$ such that for all $x,y,z \in V$ and for all $c \in F$ the following four properties are satisfied:


*

*$\langle x+y,z \rangle = \langle x,z \rangle + \langle y,z \rangle $ 

*$\langle cx,y \rangle = c\langle x,y \rangle$

*$\langle x,y \rangle = \overline {\langle y,x \rangle}$

*$\langle x,x \rangle > 0$ if $x \neq 0$


But the contraposition of property 4 is: $x = 0$ if $\langle x,x \rangle \le 0$. And there is a theorem: $\langle x,x \rangle = 0$ if and only if $x=0$.
I can understand that $\langle x,x \rangle = 0$ if $x=0$, but for $x=0$ if $\langle x,x \rangle = 0$, what if $\langle x,x \rangle < 0$? I mean $x=0$ if $\langle x,x \rangle < 0$.
I want to know


*

*Is there any $x$ that satisfies $\langle x,x \rangle < 0$?

*Or if it does not exist, how can I prove that $\langle x,x \rangle < 0$ doesn't exist?

 A: Point 4 says: if $x\neq 0$, then $\langle x,x\rangle > 0$. As you observe, $\langle x,x\rangle = 0$ iff $x = 0$. So this immediately answers your question 1: Break into two cases, $x\neq 0$ and $x=0$. In the first case, $\langle x,x\rangle > 0$. In the second case, $\langle x,x\rangle = 0$. In neither case do we have $\langle x,x\rangle < 0$, and the cases exhausted the vector space, so we have your desired result.
(Note that if we remove the fourth hypothesis, we generalize to what are called indefinite inner products. These are important in the theory of relativity. See more on Wikipedia.)
A: Assume you have $x\in V$ such that $\langle x,x\rangle \leq 0$. We will prove that then we must have both $\langle x,x\rangle = 0$ and $x=0_V$ (I denote the zero element of $V$ by $0_V$ to avoid confusion with the scalar $0\in\mathbb{R}$).


*

*From (the contrapositive of) 4., you know that $x=0_V$, as "if $x\neq 0_V$ then $\langle x,x\rangle > 0$."

*But $0_v = c\cdot 0_v$ for $c=0$, so applying 2. we get 
$$
\langle x,x\rangle = \langle cx,x\rangle = c\langle x,x\rangle = 0\langle x,x\rangle = 0
$$
and therefore $\langle x,x\rangle=0$.
To summarize, for any $x\in V$ we have that $\langle x,x\rangle \leq 0$ implies ($x=0_V$ and $\langle x,x\rangle =0$). So there is no $x\in V$ such that $\langle x,x\rangle < 0$.
