Functional on $V\times W$ 
Let $g:V\times W\to K$ be a bilinear map written $\langle ,\rangle$. For each $w\in W$, the map of $V$ into $K$ given by:
$L_w:v\mapsto\langle v,w\rangle$ is a functional on $V$, and that map $w\mapsto L_w$ is a linear map of $W$ into the dual space of $V^{*}$.
Show that the following conditions are equivalent:
(i) The map $L:W\to V^*$ is injective (its kernel is $\{0\}$).
(ii) If $C$ is a matrix associated with $g$, then $C$ has a rank $n$, where $n=\dim W$.

(i) $w\mapsto L_w$ is an injective map iff $\ker L=0$
By non degeneracy property
If $v\in V$ and $v\neq 0$.
$L_w(v)=\langle v,w\rangle=0$ iff $w=0$ $\implies \ker(L_w)=0$.
If we consider $\mathscr{B},\mathscr{B}^*$ to be the bases of $V$ and $V^*$
$c_{ij}=g(\mathscr{B},\mathscr{B}^*)=\langle \mathscr{B},\mathscr{B}^*\rangle$
Therefore $c_{ij}$ gives rise to a diagonal matrix of dimension $n$.
By a previous theorem it is proven $\dim V=\dim V^*\implies \dim V^*=n$
As $L:W\to V*$ is injective then $\dim W=n$.
Questions:
1) How can we know "$L:W\to V^*$ is injective then $\dim W=n$"? We need an isomorphism right? However (I) does not require an isomorphism.
2) Is my proof right? I feel there are steps not well justified as the last one.
3) Is Matrix C a diagonal matrix?
4) Can someone help me correct or provide me an alternative proof?
Thanks in advance!
 A: Actually the map $L$ is not only injective, but also surjective.
Proof:
Assume that $\exists w, w' \in W$ such that $L(w) = L(w')$, i.e $L_w = L{w'}$, so 
$$\forall v \in V \qquad <v, w> = <v, w'> \Rightarrow <v, w - w'> = 0$$, and by the non-degenerate character of $<,>$, $w=w'$, so $L$ is injective.
And as for your second question, since we have a non-degenerate bilinear function between $V$ and $W$, $V $ and $W$ are dual spaces(see the side note at the end of the answer), so 
$$dim V^* = dim V = dim W$$
In another way, you can also say that, since $L$ is injective
$$dim W \leq V^*.$$
So now consider the other way around, i.e let $L' : V \to W^*$,
since this map will also be injective, we have 
$$dim V \leq W^*$$
So, combining two result with the fact that $dim V = dim V^*$, we get
$$dim V = dim W$$
A side note, the definition of dual space of a vector space is more general than just the space of linear functionals of that vector space $L(V)$.In fact, the exact definition of a dual vector space of $V$ is a vector space $W$ such that there is a non-degenerate bilinear function defined between $V$ and $W$.
A: $$
\newcommand{\bv}{{\mathbf v}}
\newcommand{\bw}{{\mathbf w}}
\newcommand{\bC}{{\mathbf C}}
\newcommand{\bz}{{\mathbf 0}}
\newcommand{\bx}{{\mathbf x}}
$$
Your proof is wrong. It assumes non-degeneracy, which is not given in the hypotheses of the problem. It assumes $V$ is finite dimensional, which is not given in the hypotheses. It refers to "the bases" of $V$ and $V*$, but unless $V$ has dimension zero, or the underlying field is very small (like $F_2$), there are many possible bases for $V$ and $V*$. 
As for providing an alternative proof...I might consider providing one, but only once you answer the clarifying questions I've asked. 

Suppose that $L$ is injective. 
Pick bases $\bw_1, \ldots, \bw_n$ of $W$, and a basis $\bv_1, \ldots, \bv_k$ of $V$ (which we can do because both are finite-dimensional). 
Let $c_{ij} = g(\bv_i, \bw_j)$ for $i = 1, \ldots, k; j = 1, \ldots, n$. 
Suppose that the rank of $C$ is some number $p$ less than $n$; then there is some nontrivial linear combination of the $n$ columns of $c$ whose value is zero, i.e., there are constants $s_1, \ldots, s_n$, not all zero, with 
$$
s_1 \bC_1 + s_2 \bC_2 + \ldots + s_n \bC_n = \bz,
$$
where $C_i$ denotes the $i$th column of $C$. The $j$th row of the equation above says that 
$$
s_1 c_{j1} + \ldots + s_n c_{jn} = 0.
$$
This must hold for $j = 1, \ldots, k$. 
Now substituting the definition of $c_{ij}$, we get
$$
s_1 g(\bv_j, \bw_1) + \ldots + s_n g(\bv_j, \bw_n) = 0,
$$
which, by bilinearity of $g$, gives 
$$
g(\bv_j, s_1 \bw_1 + \ldots + s_n \bw_n) = 0
$$
for all $j$. 
Letting $\bx = s_1 \bw_1 + \ldots + s_n \bw_n$, which is nonzero because the $\bw$s form a basis and not all $s_j$ are zero, we have
$$
g(\bv_j, \bx) = 0
$$
for all $j$, Hence for any coefficients $t_i$, we have
$$
g(\sum_i t_j \bv_j , \bx) = 0
$$
and since the $\bv$s are a basis for $V$, this shows that $g(., \bx)$ is everywhere zero on $V$, or to say it differently, shows that $L_\bx$ is the $\bz$ element of $V^{*}$, contradicting the claim that $L$ is injective. 

Now forget all the symbols used in the previous part except for the bases $\{\bv_i\}$ and $\{\bw_j\}$.
Now suppose that $L$ is not injective. I'll show that the rank of the matrix is less than $n$. 
Because $L$ is not injective, we have $L_{\mathbf a} = L_{\mathbf b}$ for some distinct $\mathbf a, \mathbf b$. Let $\bx = \mathbf b - \mathbf a$. Then $L_{\mathbf x} = L_{\mathbf b} - L_{\mathbf a} = 0$. Write $\bx$ as a linear combination of the elements of our basis for $W$: 
$$
\bx = s_1 \bw_1 + \ldots + s_k \bw_k,
$$
and note that not all the $s_i$ are zero (for if they were, $\bx$ would be zero, which is not, because $\mathbf a \ne \mathbf b$). 
For every $\bv \in V$, we have $L_\bx(\bv) = 0$. In particular, for $\bv_1, \bv_2, \ldots$, we have $L_\bx(\bv_i) = 0$, whence $g(\bv_i, \bx) = 0$. 
Following the same reasoning as before, but in reverse, we get, for $j = 1, \ldots, k$, that
$$
s_1 g(\bv_j, \bw_1) + \ldots + s_n g(\bv_j, \bw_n) = \bz,
$$
and hence
$$
s_1 \bC_1 + \ldots + s_n \bC_n = \bz,
$$
where $\bC_i$ denotes the $i$th column of the matrix $C$. This is a nontrivial relation among the $n$ columns of $C$, hence the set of columns cannot be linearly independent, hence the rank of $C$ (the largest number of linearly independent columns) cannot be $n$. 

Note that this proof does not rely on $V$ being finite dimensional, i.e., every step works just fine even if $k = \infty$. It only (as written) requires that $V$ have a countable basis. I suspect it could be adjusted slightly to handle the case where $V$ has an "uncountable basis", but then you get into messy details about just what that term might mean, and I don't want to do that. 
