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?

  • 1
    $\begingroup$ It's actually the "positive definite implies non-degenerate" part that is trivial and "non-degenerate implies positive definite" which is non-trivial. $\endgroup$
    – Bruno B
    Commented Feb 11, 2023 at 11:04
  • $\begingroup$ @BrunoB Thanks. Can you please refer to some results of the proof? Thank you! $\endgroup$
    – Hermi
    Commented Feb 11, 2023 at 11:07
  • $\begingroup$ Have you seen the Cauchy-Schwarz inequality yet? $\endgroup$
    – Bruno B
    Commented Feb 11, 2023 at 11:27
  • $\begingroup$ @BrunoB Yes, $|<u,v>|\le \|u\|\|v\|$. $\endgroup$
    – Hermi
    Commented Feb 11, 2023 at 21:33
  • $\begingroup$ @BrunoB So how to prove it based on C-S inequality? $\endgroup$
    – Hermi
    Commented Feb 11, 2023 at 21:34

1 Answer 1


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.


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