Rearrange equation involving matrix multiplication and Hadamard product I have the following equation:
$Y = [ C \bullet R ] \times X \times[C^T \bullet R^T ]$
In my textbook, this conveniently rearranges to:
$Y = [ C \times X \times C^T ] \bullet [R \bullet R^T ]$
where $\times$ denotes matrix multiplication, $\bullet$ is Hadamard (pointwise multiplication), and $R^T$ is the transposition of $R$.
How do they manage to rearrange this?
I've played around with it, and it seems that they're using the following identity:
$[ A \bullet B ] \times C = A \bullet [B \times C]$
I've tried proving it true, without success.  For example, if
$D = [ A \bullet B ] \times C$ and $E = A \bullet [B \times C]$,
then
$D_{00} = \sum_{k=0}^m A_{0k} B_{0k}C_{k0}$
$E_{00} = A_{00} \sum_{k=0}^m B_{0k}C_{k0}$
As $D_{00} \neq E_{00}$, the identity is false.
Where have I gone wrong?
EDIT
I'm working with $4 \times 4$ matrices here.
EDIT 2
Here are the matrices I'm dealing with:
$
C = 
\begin{pmatrix}
1 &  1  &  1 &  1\\\\
2 &  1  & -1 & -2\\\\
1 & -1  & -1 &  1\\\\
1 & -2  &  2 & -1\\\\
\end{pmatrix}
$
$
R =
\begin{pmatrix}
\frac{1}{2}         & \frac{1}{2}         & \frac{1}{2}         & \frac{1}{2}\\\\
\frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}}\\\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\\
\frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}} & \frac{1}{\sqrt{10}}\\\\
\end{pmatrix}
$
$X$ is an arbitrary matrix with integer-only values.
 A: With these matrices the identity is true.
In order to prove it, notice that $R=DE$, where $D=\operatorname{diag}(\frac12,\frac1{\sqrt{10}},\frac12,\frac1{\sqrt{10}})$ and $E$ is the matrix of all ones. You can juggle around the diagonal factor $D$ (your proof shows that the "triple product identity" holds if $A$ is diagonal) and in particular you can reabsorb it into $C$, which is particularly convenient since now $Q=DC$ is orthogonal (that's where all those $\sqrt{10}$ come from, I guess). 
So after moving the diagonal factors the two terms become
$$
Y_1=(Q \bullet E)\times X\times (Q^T\bullet E^T)
$$
$$
Y_2=(Q \times X \times Q^T) \bullet (E\bullet E^T)
$$
Here all the remaining Hadamard products have $E$ as one factor and thus are trivial.
Anyway, it definitely looks like someone is overcomplicating things in that book; they could have removed those $R$s and those Hadamard products completely: you can replace every $R\bullet M$ with $D\times M$ and every $M\bullet R^T$ with $M\times D$.
A: What is true, though, is that the diagonal entries of $(A \circ B)C^{T}$ and $(A \circ C)B^{T}$ coincide. (See Lemma 5.1.4 on p. 305 of Topics in Matrix Analysis by Horn and Johnson).
