Show that $\langle x, x \rangle = 0$ if and only if $x = 0$.

Question

I am trying to prove all of the properties of inner product spaces listed on Wikipedia (using the axioms of an inner product space listed on the same Wikipedia page), and I am stuck on proving the third one:

• $\langle x,x\rangle = 0$ if and only if $x = \mathbf{0}$.

There are two directions to this proof, and I have already proved the first:

• Suppose $\langle x, x \rangle = 0$. By the contrapositive of the positive definiteness property of inner product spaces, if $\langle x, x \rangle \leq 0$, then $x = \mathbf{0}$, so it follows immediately that $x = \mathbf{0}$.

But I don't know how to prove the other direction: how do we prove that $x = \mathbf{0}$ implies that $\langle x, x \rangle = 0$?

Answer 1

Using linearity and since $0 + 0 = 0$ : $$\langle 0,x\rangle = \langle 0 + 0,x \rangle = \langle 0,x\rangle + \langle 0,x \rangle \Rightarrow \langle 0,x \rangle = 0 \ \forall x$$ and in particular, for $x = 0$.

Answer 2

If $x=0$, then $x=0v$ for every $v\in V$. Thus, $\langle x, x\rangle=\langle 0v, x\rangle=0\langle v, x\rangle=0$.