Represent in a matrix form: $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ How can I represent this in a matrix form:
$\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ 
where all the entries are real and $W$ is  a known(constant) matrix and $F$ is a rectangular matrix. When I say matrix form, the matrix trace function can be used as well as part of the simplification. Essentially, the index $l$ is choosing a column in $F$ and then the summation is being done and so-forth. 
 A: Let $e_i$ denotes the $i$-th vector in the canonical basis and $u=\sum_ie_i=(1,1,\ldots,1)^\top$. Then your sum is equal to
$$
u^\top \left[\left(\sum_i e_ie_i^\top FF^\top e_iu^\top
+ \sum_j ue_j^\top FF^\top e_je_j^\top
- 2FF^\top\right) \circ W\right] u,\tag{1}
$$
where the symbol "$\circ$" denotes a Hadamard product. Since
$$
u^\top (A\circ B)u = \operatorname{trace}(B^\top A) = \operatorname{vec}(B)^\top\operatorname{vec}(A),
$$
you may also express your sum as
$$
\operatorname{trace}\left(W^\top \left(\sum_i e_ie_i^\top FF^\top e_iu^\top
+ \sum_j ue_j^\top FF^\top e_je_j^\top
- 2FF^\top\right) \right)\tag{2}
$$
or
$$
\operatorname{vec}(W)^\top
\left[\sum_i (ue_i^\top)\otimes(e_ie_i^\top) + \sum_j(e_je_j^\top)\otimes(ue_j^\top) - 2I\otimes I\right]
\operatorname{vec}(FF^\top),\tag{3}
$$
where "$\otimes$" denotes Kronecker product (tensor product). Either way, I think $\sum_{ijl}(F_{il}-F_{jl})^2W_{ij}$ is a much simpler expression.
A: Let $E$ be a matrix with the same size as $F$, and with all elements $E_{jl}=1$. 
Then there are several different ways to write the expression using the Hadamard ($A\!\circ\!B$), Matrix ($AB$), and Frobenius ($A\!:\!B$) products.
$$\eqalign{
  S &= (F\circ F):WE + (F\circ F):W^TE - 2\,FF^T:W \cr 
    &= (F\circ F):(W+W^T)E - 2\,FF^T:W \cr 
    &= (\,(F\circ F)E^T-FF^T):(W+W^T) \cr 
    &= (\,(F\circ F)E^T-FF^T):K \cr 
}$$ 
where $K = W+W^T$
We can also find the differential of that final expression as 
$$\eqalign{
 dS &= (\,(dF\circ F)E^T+(F\circ dF)E^T-dFF^T-FdF^T):K \cr 
    &= 2\,(dF\circ F):KE-2\,dF:KF \cr 
    &= 2\,dF:F\circ(KE)-2\,dF:KF \cr 
    &= (2\,F\circ(KE)-2\,KF):dF \cr 
}$$
And since $dS=\frac{\partial S}{\partial F}:dF\,\,$ the derivative must be
$$\eqalign{
 \frac{\partial S}{\partial F} &= 2\,F\circ(KE)-2\,KF \cr 
}$$
A: Consider a matrix with components given by
$$D_{ij} = \sum_\ell (F_{i\ell}-F_{j\ell})^2$$
This is the Euclidean distance matrix for the rows of the matrix $F$ (or equivalently the columns of $F^T$).
There are many different expressions for this matrix. My personal favorite is
$$D = (F\odot F)J^T+J(F\odot F)^T - 2FF^T$$
where $J$ is the all-ones matrix the same size as $F$, and $\odot$ denotes the elementwise/Hadamard product.
So by choosing to sum over the $\ell$-index first, the current problem can be greatly simplified
$$
\sum_{ij}W_{ij}\sum_{\ell}(F_{i\ell}-F_{j\ell})^2
= \sum_{ij} W_{ij} D_{ij}
 = {\rm Tr}(WD)
$$

UPDATE
If the repeated $(i,j)$ indexes were not meant to imply summation (i.e the Einstein convention), then the expression in question is equal to the Hadamard product $(W\circ D)$ since
$$
W_{ij}\sum_{\ell}(F_{i\ell}-F_{j\ell})^2
\;=\; W_{ij} D_{ij} \;=\; (W\circ D)_{ij}
$$
