Bilinear form is non degenerate if and only if $f$ is bijective

bilinear-formlinear algebra

Let $$(V,\langle \cdot,\cdot\rangle)$$ a finite dimensional euclidean vector space and $$f\in \mathcal{L}(V,V)$$. Then defiene a bilinear form on $$V$$ $$\beta(v,w)=\langle v,f(w)\rangle$$

1. Prove that $$\beta$$ is non degenerate iff $$f$$ is bijective.

According to the definition of non degenerate (what I've learnd), if $$\beta$$ is non degenerate in first variable, when $$\forall w\in V\ \beta(v,w)=0\Rightarrow v=0$$
and if $$\beta$$ is non degeneratte in second variable, when $$\forall v\in V\ \beta(v,w)=0\Rightarrow w=0$$

But how can I get the bijectivity of $$f$$ with this definition?
$$\forall v\in V\ \beta(v,w)=\langle v,f(w)\rangle\ \Rightarrow\ w=0,f(w)=0$$ Besides I tried to prove f is injective or surjective, since the linear function f is totally in $$V$$, then $$f$$ is bijective. Could you please give me some hints, thank you!

$$(f:\text{not invertible} \iff\beta: \text{degenerate}) \equiv (f:\text{invertible} \iff\beta: \text{non-degenerate})$$

In the 1st argument

• $$\beta: \text{degenerate} \implies f:\text{not invertible}$$

If $$\beta$$ is degenerate then there is a nonzero vector $$v \in V$$ such that for all $$x \in V$$ $$0 = \beta(v,x) = \langle v, f(x)\rangle \implies \text{Im}f \subset \{v\}^\perp \subset V,$$ so the image is proper subspace and $$f$$ is not surjective.

• $$f:\text{not invertible} \implies \beta: \text{degenerate}$$

If $$f$$ is not invertible then its image is proper subspace $$\text{Im} f \subsetneq V$$, and wlog non trivial, then its orthogonal complement $$W=(\text{Im} f)^\perp$$ is nonempty. Take any $$v \in W$$ then for all $$y \in V$$ $$\beta(v,y) := \langle v, f(y) \rangle = 0.$$

In the 2nd Argument

• $$\beta: \text{degenerate} \implies f:\text{not invertible}$$

If $$\beta$$ is degenerate then there is a nonzero vector $$x \in V$$ such that for all $$w \in V$$ $$0 = \beta(w,x) = \langle w, f(x)\rangle \implies f(x) = 0,$$ by the non-degeneracy of the inner product. It follows that the kernel is nontrivial and $$f$$ is not injective.

• $$f:\text{not invertible} \implies \beta: \text{degenerate}$$

Its kernel is not trivial and so there is a nonzero vector $$y \in \ker f$$ such that for all $$v \in V$$ the bilinear form $$\beta(v,y) := \langle v, f(y) \rangle = \langle v, 0 \rangle = 0$$ is degenerate.