$\varphi(f)$ is invertible iff $f$ is non-degenerate? Let $E$ be the vectorial space of the bilinear functions $\varphi: \mathbb R^n\times \mathbb R^n\to \mathbb R$. 
Then, there is a canonical isomorphism between $E$ and the set of the real matrices $n\times n$, $M(n\times n)$: 
$$\varphi:E\to M(n\times n), \ \varphi(f)=f(e_i,e_j)_{ij}$$
My question is can we say that $\varphi(f)$ is invertible iff $f$ is non-degenerate?
Definition: 
$f$ is non-degenerate if $f(x,y)=0$ for every $y\in \mathbb R^n\implies x=0$
Thanks
 A: Any $f$ determines a linear mapping $F \colon \mathbf{R}^n \to (\mathbf{R}^n)^{*}$ from the vector space to its dual, defined by $F(x)(y) = f(x,y)$. (The dual of a vector space $V$ is the space $V^{*}$ of linear mappings from $V$ to $\mathbf{R}$.)
The condition that $f$ is non-degenerate means precisely that $F$ has trivial kernel, or, equivalently, is injective. Since the vector spaces $\mathbf{R}^n$ and $(\mathbf{R}^n)^{*}$ involved both have same finite dimension $n$, this occurs if and only if $F$ is bijective.
It turns out that matrix of $F$ is the transpose of $\varphi(f)$. Thus $\varphi(f)$ is invertible if and only if $f$ is non-degenerate.
EDIT: I will phrase the same proof in a more concrete way. 
Let $M$ be the matrix $\varphi(f)$, and assume you've written $x$ and $y$ as column vectors $X$ and $Y$. Then $f(x,y) = X^T M Y$. For fixed $x$, the condition that $f(x,y)= 0$ for ever $y$ means that $X^T M Y = 0$ for every column vector $Y$. This is equivalent to the row vector $X^T M$ being zero, which, after transposition, is equivalent to the column vector $M^t X$ being zero. 
Thus the condition that $\varphi$ is non-degenerate amounts to asking that $M^T X = 0$ only when $X = 0$. This means that the linear mapping corresponding to $M^T$ is injective. Since $M$ is square, this is in turn equivalent to $M$ being invertible.  
