Hessian wrt. MATRIX-VARIABLE for a Quadratic Inner Product. Given standard matrix inner product,
\begin{equation}
\begin{aligned}
f(\textbf{X}) := &   \;\;\;\;  {\langle}{\textbf{X}, \textbf{A}\textbf{X}}{\rangle}\\
=& \;  \text{tr} (\textbf{X}^{\text{T}} \textbf{A}\textbf{X}),\\
& \textbf{X} \in \mathcal{R}^{n \times r}, \text{ the variable,}\\
& \textbf{A} \in \mathcal{R}^{n \times n}, \text{ a constant matrix
and isn't necessarily be symmetric}.
\end{aligned}
\end{equation}
I want to calculate the Hessian with respect to ${\textbf{X}}$
which is a matrix not a vector. I know how to compute the gradient
which is,
\begin{equation}
\begin{aligned}
&   \nabla_\textbf{X} f(\textbf{X})=
\begin{cases}
{\textbf{2AX}, \; \; \; \; \; \; \; \; \;
\text{ if } \textbf{A} = \textbf{A}^{\text{T}}},\\
{\textbf{(A+A}^{\text{T}}) \textbf{X}, \text{ else. }}
\end{cases}\\
& \in \mathcal{R}^{n \times r}
\end{aligned}
\end{equation}
but it is highly unclear how I do for a matrix variable. And in general, it is hard to find a material available and clear for this. Of course, the definition is in Wikipedia but for a matrix, I need a small example like
$\textbf{X} \in \mathcal{R}^{3 \times 2}, \;
\textbf{A} \in \mathcal{R}^{3 \times 3},$
then it will become clear.
In this case of the small example, the dimension of the Hessian Matrix will become $\textbf{X} \in \mathcal{R}^{6 \times 6}$ as far as believe. 
And hopefully, there will exist 
$\textbf{the neat mathematical expression to denote 
the resulting hessian matrix}$
for this function as it does for the gradient.
$\textbf{With a clear example please}$, thanks in advance.
It will definitely help many people because this is fundamental but not well accessible.
 A: Since you know how to calculate the gradient, let's start by taking the differential of that 
$$\eqalign{
 S &= A+A^T \cr
 G &= \nabla f = SX \cr
dG &= S\,dX \cr 
}$$
There are two ways to proceed: vectorize the equation or use tensors.
Vectorization flattens the $(dG,dX)$ matrices into vectors and the Hessian into a matrix.
$$\eqalign{
dg &= (I\otimes S)\,dx \cr
H = \frac{\partial g}{\partial x} &= I\otimes S \cr
}$$
where $\otimes$ represents the Kronecker product and $\,\,dx={\rm vec}(dX)$.
But the true Hessian is a fourth-order tensor.
$$\eqalign{
dG &= S{\mathcal E}:dX \cr
{\mathcal H} = \frac{\partial G}{\partial X} &= S{\mathcal E} \cr
}$$
where ${\mathcal E}$ is a tensor constant whose components can be written in terms of Kronecker deltas 
$$\eqalign{
{\mathcal E}_{ijkl} &= \delta_{ik}\delta_{jl} \cr
}$$
The colon represents the double-contraction product
$$B={\mathcal E}:X \implies B_{ij}=\sum_k\sum_l {\mathcal E}_{ijkl}\,X_{kl}$$
while juxtaposition represents the single-contraction product.
The components of the Hessian are equal to 
$$
{\mathcal H}_{ijkl}
 = \frac{\partial G_{ij}}{\partial X_{kl}}
 = \sum_nS_{in}{\mathcal E}_{njkl}
 = S_{ik}\delta_{jl}
$$
