What is the relation between a matrix as a linear function versus the same matrix as a bilinear function? Given an $n \times n$ matrix $A$, we can define a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by $T(x)=Ax$.
We could also define a bilnear function $T: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ by $T(x,y) = y^TAx$.
Is there a relation between these two uses of the matrix?
Also, we could do the same thing with an $n \times m$ matrix and get a linear function $\mathbb{R}^m \rightarrow \mathbb{R}^n$ and a bilinear function $\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}$. Is the relationship between using $A$ to define a linear function versus a bilinear function the same in this case?
 A: Any bilinear form $b\colon \mathbb R^n\times\mathbb R^n\to\mathbb R$ naturally corresponds to a linear map $\Phi_b\colon \mathbb R^n\to(\mathbb R^n)^*$, where $(\mathbb R^n)^*=\operatorname{Hom}(\mathbb R^n,\mathbb R)$ is the dual space. This correspondence is given by
\begin{align}
b \quad\longmapsto\quad \Phi_b\colon\mathbb R^n&\to(\mathbb R^n)^*,\\
x &\mapsto b(x,-).
\end{align}
Here $b(x,-)$ denotes the linear map $\mathbb R^n\to\mathbb R$ sending $y$ to $b(x,y)$.
Considering the elements of $\mathbb R^n$ as column vectors, we can consider the elements of $(\mathbb R^n)^*$ as row vectors: for $\rho\in(\mathbb R^n)^*$ and $x=(x_1,\dots,x_n)^T$ we have
\begin{align*}
\rho(x) &= \rho(x_1 e_1+\cdots+x_n e_n) = x_1 \rho(e_1) +\cdots + x_n \rho(e_n)
\\&= \begin{pmatrix} \rho(e_1) & \rho(e_2) & \dots & \rho(e_n)\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}.
\end{align*}
Using this identification, there is an isomorphism ${}^T\colon(\mathbb R^n)^*\to\mathbb R^n$ given by $y\mapsto y^T$.
Now starting with $b(x,y) = y^TAx$ this yields $\Phi_b(x)=(y\mapsto y^TAx)$ which we identified with the row vector $\Phi_b(x)=(Ax)^T$. Using the above isomorphism we then obtain a linear map $\mathbb R^n\to\mathbb R^n$:
\begin{align}
\mathbb R^n &\xrightarrow{\Phi_b}\ (\mathbb R^n)^*\ \xrightarrow{T} \mathbb R^n,\\
x &\longmapsto (Ax)^T \mapsto Ax.
\end{align}
This is how starting with the bilinear map defined by $A$ we obtain the linear map defined by $A$. You can of course go in the other way as well.

More abstractly you can think of this in terms of the tensor-hom adjunction together with the (basis dependent!) isomorphism $(\mathbb R^n)^*\cong(\mathbb R^n)$:
\begin{align}
\{\text{bilinear maps $\mathbb R^n\times\mathbb R^n\to\mathbb R$}\}
&\leftrightarrow
\operatorname{Hom}(\mathbb R^n\otimes\mathbb R^n,\mathbb R)
\\&\cong
\operatorname{Hom}(\mathbb R^n,\operatorname{Hom}(\mathbb R^n\to\mathbb R))
\\&=
\operatorname{Hom}(\mathbb R^n,(\mathbb R^n)^*)
\\&\cong
\operatorname{Hom}(\mathbb R^n,\mathbb R^n).
\end{align}
A: Redefining symbols to avoid ambiguity: $T: \mathbb{R}^n \to \mathbb{R}^n$ is the linear map defined as $T(x) = Ax$ and $S: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ is the bilinear map defined as $S(x, y) = y^T A x$.

Constructing bilinear functions from linear functions using inner product
One way to understand $S$ is as composition of $T$ with the standard inner product $\phi: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ defined as $\phi(x, y) = y^T x$, namely
$$
S(x, y) = \phi(T(x), y).
$$
This view allows us to notice some properties of $S$ based on properties of $T$ and known properties of $\phi$. For example, since $\phi$ is known to be non-degenerate, $S$ is non-degenerate if and only if $T$ is an isomorphism.
The construction is readily generalized to the $n \times m$ case by composing $T: \mathbb{R}^m \to \mathbb{R}^n$ with $\phi$.

Change of basis
For a fixed matrix $A \in \mathbb{R}^{n \times n}$ the construction above yields two functions: a linear function $T_A: \mathbb{R}^n \to \mathbb{R}^n$ and a bilinear function $S_A: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$. The construction takes place in a fixed basis, but the resulting functions $S$ and $T$ are independent of basis, so it is natural to ask how their matrix representation changes under basis transformations.
It easy to see that the matrix representing a linear function transforms differently than the matrix representing a bilinear function. Let $B$ denote an invertible matrix describing a change of basis. Then
$$
T_A = T_{A'}
$$
whenever $A' = B^{-1}AB$, i.e. the matrices representing a fixed linear map are similar. On the other hand,
$$
S_A = S_{A^{''}}
$$
whenever $A^{''} = B^TAB$, i.e. the matrices representing a fixed bilinear map are congruent.
This shows that care must be taken when using matrix representations of linear and bilinear functions. Even when a linear function $T$ and a bilinear function $S$ are represented by the same matrix in one basis, it does not imply that they are represented by the same matrix in another basis (unless the basis transformation is orthogonal).
A: Given
$$
{\bf y}_{\,1}
  = \left( {\matrix{ {y_{1,1} }  \cr  \vdots \cr {y_{h,1} } \cr } } \right)
 = {\bf A}\,{\bf x}_{\,1} 
$$
then
$$
{\bf Y}
 = \left( {\matrix{ {y_{1,1} } &  \cdots  & {y_{1,h} }  \cr \vdots  &  \ddots  &  \vdots   \cr 
   {y_{h,1} } &  \cdots  & {y_{h,h} }  \cr  } } \right)
 = {\bf A}\,{\bf X}
$$
Therefore
$$
{\bf Y}^{\,T} {\bf Y} = {\bf Y}^{\,T} {\bf A}\,{\bf X}
 = {\bf X}^{\,T} {\bf A}^{\,T} {\bf A}\,{\bf X}
 = \left( {\matrix{  {{\bf y}_{\,1}  \cdot {\bf y}_{\,1} } &  \cdots
  & {{\bf y}_{\,1}  \cdot {\bf y}_{\,h} }  \cr     \vdots  &  \ddots  &  \vdots   \cr 
   {{\bf y}_{\,h}  \cdot {\bf y}_{\,1} } &  \cdots  & {{\bf y}_{\,h}  \cdot {\bf y}_{\,h} }  \cr 
 } } \right)
$$
