How do I solve for a matrix in this linear algebra expression? In the problem I am working on I have an expression that looks like
$$B^T\operatorname{diag}(Bx) y$$
where $B$ is a known $m \times n$ matrix, $x$ is an unknown $n \times 1$ vector of variables, and $y$ is a known $m \times 1$ vector, all of real numbers.
The expression above can always be rewritten in the form $Px$ by expanding and collecting like terms, where $P$ is an $n \times n$ matrix. I would like to know how to calculate $P$ using only matrix and vector products.
As an example, take $B =  \left( \begin{array}{ccc}
-1 & 1 & 0 \\
-1 & 0 & 1\end{array} \right), y = \left( \begin{array}{c}
2 \\
3\end{array} \right) $.
Then
\begin{align}
B^T \operatorname{diag}(Bx) y & = \left( \begin{array}{cc}
-1 & -1 \\
1 & 0 \\
0 & 1\end{array} \right) \left( \begin{array}{ccc}
-x_1 + x_2 & 0\\
0 & -x_1 + x_3\end{array} \right) \left( \begin{array}{c}
2 \\
3\end{array} \right) \\[10pt]
& = \left( \begin{array}{c}
-5x_1 -2x_2 -3x_3 \\
-2x_1 + 2x_2 \\
-3x_1 + 3x_3\end{array} \right) = \left( \begin{array}{cc}
-5 & -2 & -3 \\
-2 & 2 & 0 \\
-3 & 0 & 3\end{array} \right) \left( \begin{array}{c}
x_1 \\
x_2 \\
x_3\end{array} \right)
\end{align}
Therefore $P = \left( \begin{array}{cc}
-5 & -2 & -3 \\
-2 & 2 & 0 \\
-3 & 0 & 3\end{array} \right)$
But I do not know how to calculate $P$ as an expression of the other known quantities, $P = F(B,y)$. Thanks.

Note that $x \mapsto B^T \operatorname{diag}(Bx)y$ is a linear transformation on $x$.  To find the matrix of this transformation (that is, the matrix $P$ for which $Px = B^T \operatorname{diag}(Bx)y$), we can see what happens to the standard basis vectors in order to get the columns of $P$.
Long story short: let $B_k$ denote the $k$th column of $B$.  Let $P_k = B^T \operatorname{diag}(B_k)y$.  We then have
$$
P = [P_1 \quad P_2 \quad \cdots  \quad P_n]
$$
If we use $\circ$ to denote the Hadamard product, then we could also say 
$$
P_k = B^T(B_k \circ y) = B^T(y \circ B_k) = B^T \operatorname{diag}(y) B_k
$$
From which one may deduce that
$$
P = B^T \operatorname{diag}(y) B
$$
(as inspired by the other answer)
