Bilinear form question to prove isometry? I am given $f$ to be a bilinear form on $V$, a finite dimensional vector space. It is symmetric with $f(v,w)=f(w,v)$ for all $v,w \in V$. Also $f(v,w) = 0 \forall w \in V$ implies that $v=0$. If I define the map $L_w(v) = f(v,w)$, then I need to show that $L: V \rightarrow V^*$ is an isomorphism. ($V^*$ is defined as dual of $V$ so its the set of all bilinear forms on $V$)
I was thinking to show this, we start with defining set $B = \{v_1, .. v_n \}$ such that it is a basis for $V$. Then I show that $B_L = \{L_w(v_1), ... L_w(v_n) \}$ forms a basis for $V^*$. Is this enough? I am given a hint of using rank nullity?
Do I need to show there exists an inverse such that $L^{-1}L = $ identity?
Please help any advice or push in the right direction would be much appreciated!
 A: Yes, it is enough because of
Lemma 1: A linear map $\;f: V\to W\;$ of vector spaces over the same field, is injective iff 
$$\forall\;\text{lin. ind. set}\;\{v_1,...,v_m\}\subset V\;,\;\;\text{also}\;\{fv_1,...,fv_m\}\;\text{is lin. ind. in}\;W$$
Lemma 2: With the given data as above, if also $\;\dim V=\dim W<\infty\;$ , then $\;f\;$ is an isomorphism iff it is injective
A: Let $V$ be a finite dim. vector space on $\mathbb R$ (for simplicity). Let $\{e_i\}$ be a basis of $V$ s.t. $f(e_i,e_j)=\delta_{ij}$, i.e. it is orthonormal w.r.t. $f$ (such basis can be always found). The map
$$L: V\rightarrow V^{*} $$
is an isomorphism because $\operatorname{ker}(L)=\{0\}$ and it is surjective.
The first condition follows from non degeneracy of $f$:
$$\operatorname{ker}(L)=\{v\in V: L_v=0 \in V^{*}\}=
\{v\in V: L_v(w)=f(v,w)=0~\forall w\in V\}=\{0\}.$$
Surjectivity can be proven by writing $\varphi=\sum_i\varphi_ie^{*}_i\in V^{*}$ ($\{e^{*}_i\}$ is the dual basis in $V^{*}$) and selecting the map $L_{v}$, with 
$v=\sum_j \varphi_je_j$. It follows that
$$L_v(e_k)=\sum_j\varphi_jf(e_k,e_j)=\varphi_k=\varphi(e_k), $$
for all $e_k$ in the basis $\{e_i\}$ of $V$.
