# Show that a positive symmetric bilinear form is positive definite (i.e., a inner product) if and only if it is nondegenerate.

Show that a positive symmetric bilinear form $$\langle\cdot, \cdot \rangle$$ on a real vector space $$V$$ with $$dim V=n<\infty$$ is positive definite (i.e., a inner product) if and only if it is nondegenerate.

Here a symmetric bilinear form $$\langle\cdot, \cdot \rangle$$ is called nondegenerate if the only $$v\in V$$ so that $$\langle v, w \rangle=0$$ for all $$w\in V$$ is that $$v=0$$.

My proof: if $$\langle\cdot, \cdot \rangle$$ is nondegenerate, then it is enough to show that $$\langle\cdot, \cdot \rangle$$ is Definiteness (i.e., $$\langle v, v \rangle=0$$ iff $$v=0$$). This is trivial.

But how to show the inverse statement?

• It's actually the "positive definite implies non-degenerate" part that is trivial and "non-degenerate implies positive definite" which is non-trivial. Feb 11 at 11:04
• @BrunoB Thanks. Can you please refer to some results of the proof? Thank you! Feb 11 at 11:07
• Have you seen the Cauchy-Schwarz inequality yet? Feb 11 at 11:27
• @BrunoB Yes, $|<u,v>|\le \|u\|\|v\|$. Feb 11 at 21:33
• @BrunoB So how to prove it based on C-S inequality? Feb 11 at 21:34

Going to detail what I said in my comments:

You seem to be confused about which implication is trivial or not.
It is true that for a fixed $$v$$, the property "$$(A):\,\,\, \forall w \in V,\, \langle v,w\rangle = 0$$" implies that $$\langle v,v\rangle = 0$$ by taking $$w = v$$.
However what that means is that, if $$\langle\cdot,\cdot\rangle$$ is positive-definite, then, given any $$v$$ satisfying $$(A)$$, we get that $$v = 0$$, which provides "positive-definite implies nondegenerate" and not the other way round!

As for the other implication, let me first detail the proof for Cauchy-Schwarz when the form isn't positive-definite:

Lemma ("generalised" Cauchy-Schwarz): Let $$\langle\cdot,\cdot\rangle : V \to \mathbb{R}$$ be a positive symmetric bilinear form on $$V$$ a $$\mathbb{R}$$-vector space.
Then, we have: $$\forall (v,w) \in V^2,\quad \langle v,w\rangle^2 \leq \langle v,v\rangle\cdot\langle w,w\rangle$$

Proof: Let $$(v,w) \in V^2$$. Denote by $$P$$ the following polynomial function: $$P : t \in \mathbb{R} \mapsto \langle v + tw,v+tw \rangle \in \mathbb{R}$$ After expanding and using the symmetry, we have: $$P(t) = \langle v,v\rangle + 2t\langle v,w\rangle + t^2 \langle w,w\rangle$$ Yet by positivity of $$\langle \cdot,\cdot \rangle$$ $$P$$ is positive.
There are two cases now:

• If $$\langle w,w\rangle \neq 0$$: then the proof is just like regular Cauchy-Schwarz. $$P$$ is a positive quadratic polynomial, thus its discriminant is nonpositive, which provides what we want.
• If $$\langle w,w\rangle = 0$$: $$P$$ is then a positive affine function, thus its leading coefficient $$\langle v,w\rangle$$ must be $$0$$, which does give what we desire since $$0^2 \leq \langle v,v\rangle \cdot 0$$.

QED (for the lemma)

Now, let's suppose our $$\langle \cdot,\cdot \rangle$$ is non-degenerate. Let $$v \in V$$ be such that $$\langle v,v\rangle = 0$$.
Then, for all $$w \in V$$, by Cauchy-Schwarz: $$0 \leq \langle v,w \rangle^2 \leq \langle v,v\rangle \cdot \langle w,w\rangle = 0$$ Thus: $$\forall w \in V,\, \langle v,w\rangle = 0$$, thus, by non-degeneracy: $$v = 0$$, and $$\langle \cdot,\cdot\rangle$$ is positive-definite, hence the equivalence by double implication.