Positive definite(ness) of the inner product?

Question

I was working on an inner product proof. From what I've learned, the positive-definite axiom of inner products involves showing that:

Given a vector space $V$ over $F$, for all $v \in V$, $\langle v, v \rangle \in R$ and $\langle v, v\rangle \geq 0$ with $\langle v, v \rangle = 0$ iff $v = 0$. Or alternatively, $\langle v, v \rangle \geq 0$ with equality iff $v = 0$.

While the other axioms are fairly easy to comprehend, I'm a little confused about this one. In particular, what does it expect one to prove?

My thoughts: First, show that $\langle v, v \rangle \in R$ and $\langle v, v \rangle \geq 0$ for all $v \in V$. Then, assume $v = 0$ and show that $\langle v, v \rangle = 0$. Then, assume $\langle v, v \rangle = 0$ and show that $v = 0$.

Is that what it requires? Or, is one supposed to assume that $v = 0$ and show that $\langle v, v \rangle \geq 0$ AND $\langle v, v \rangle = 0$, and then, assume $\langle v, v \rangle \geq 0$ and $\langle v, v \rangle = 0$ and show that $v = 0$?

After seeing some proofs, I'm still unsure. Any explanation is much appreciated.

Answer 1

The first one is what it's saying. In order to satisfy the axiom as presented, one must show:

• $\langle v, v \rangle \in \Bbb{R}$ for all $v \in V$,
• $\langle v, v \rangle \ge 0$ for all $v \in V$ (note that $\langle v, v \rangle$ can only be considered $\ge 0$ if it is a real quantity),
• $\langle v, v \rangle = 0 \implies v = \vec{0}$,
• $v = \vec{0} \implies \langle v, v \rangle = 0$, or in other words, $\langle \vec{0}, \vec{0} \rangle = 0$.

That said, two of these points can be deduced from the other axioms. Inner products must also satisfy $\langle v, w \rangle = \overline{\langle w, v\rangle}$ for all $v, w \in V$, hence $$\langle v, v \rangle = \overline{\langle v, v\rangle},$$ and the only complex numbers equal to their own conjugate are real numbers. So, we actually don't really need to prove the first point.

We also don't need to prove the last point, due to the linearity condition. We know that, for all $v, w \in V$ and $\lambda \in \Bbb{C}$, $$\langle \lambda v, w \rangle = \lambda \langle v, w \rangle.$$ In particular, if we take $v = w = \vec{0}$ and $\lambda = 0$, then we get $$\langle \vec{0}, \vec{0}\rangle = \langle 0 \cdot \vec{0}, \vec{0}\rangle = 0\langle \vec{0}, \vec{0}\rangle = 0.$$ So, you really only need to prove the second and third points, which should hopefully align with the proofs you've seen so far.