Let $K$ be a field, let $V$ be an $n$-dimensional $K$-vector space, and let $f: V \times V \to K$ be a bilinear form. The bilinear form $f$ is called non-degenerate if the mapping $$V \to \operatorname{Hom}(V,K); \ v \mapsto (w \mapsto f(v,w))$$ is an isomorphism of vector spaces. Show that the following conditions are equivalent:
(i) $f$ is non-degenerate.
(ii) If $f(v,w) = 0$ for all $w ∈ V$, then $v = 0$.
(iii) There exists a basis $v_{1},\dots,v_{n}$ of $V$ such that the associated matrix $(f(v_{i}, v_{j}))_{1 \leq i,j \leq n}$ is invertible.
(iv) For every basis $v_{1},\dots,v_{n}$ of $V$, the associated matrix $(f(v_{i}, v_{j}))_{1 \leq i,j \leq n}$ is invertible.
I wonder what properties of bilinear form, $\operatorname{Hom}$, and isomorphism that need to be used here to prove those conditions?