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 $V$ $\beta(v,w)=\langle v,f(w)\rangle$
- 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
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.
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
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.
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.