Is this expression a quadratic form I have an matrix expression that basically is of the form:
\begin{equation}
tr(B X BX )
\end{equation}
Where $B$ and $X$ and nonsquare matrices.  $B$ is $p \times n$, $X$ is $n \times p$.  
It seems to me this trace expression is a quadratic form because I got it as part of a longer matrix expression which represents the Hessian term of a Taylor series approximating a matrix function.  However, I can't see to get it into the form:
\begin{equation}
\text{vec}(X)^T [H] \text{vec}(X)
\end{equation}
Where $[H]$ is the square matrix representation of this quadratic form.  I want to get it into this form so I can get the eigenvectors of this matrix $[H]$.  $\text{vec}(X)$ is the vector of length $pn$ where the columns of $X$ are on top of each other. 
Can someone help me confirm whether this trace expression can indeed be converted into an explicit matrix representation?
I know that I can make this into:
\begin{equation}
\text{vec}(X^T)^T (B^T \otimes B) \text{vec}(X)
\end{equation}
But this is not what I want because it gives me $\text{vec}(X^T)$ on the left!  It seems so close yet so far.
Thanks.
 A: You are almost there.  Notice that the elements of $\mathrm{vec}(X^T)$ are just a reordering of $\mathrm{vec}(X)$.  The two are related by a permutation matrix that is sometimes known as a stride permutation.  If $X\in \mathbb{R}^{m\times n}$ then the stride permutation matrix $L_m^{mn}$ satisfies the equation $L_m^{mn}\mathrm{vec}(X)=\mathrm{vec}(X^T)$.  Therefore you have $\mathrm{trace}(BXBX)=\mathrm{vec}(X)((L_m^{mn})^T B\otimes B)\mathrm{vec}(X)$.  There is also another interesting way to obtain this result that will make future calculations simpler.  For matrices $A\in\mathbb{R}^{m_1\times n_1}$ and $B\in\mathbb{R}^{m_2\times n_2}$
define the box product
$A\boxtimes B\in\mathbb{R}^{(m_1m_2)\times(n_1n_2)}$
by 
$$ (A\boxtimes B)_{(i-1)m_2+j,(k-1)n_1+l} = a_{il}b_{jk} "=
(A\boxtimes B)_{(ij)(kl)}" $$
then $I_m\boxtimes I_n = L_{m}^{mn}$ and you can also write
$$\mathrm{trace}(BXBX)=\mathrm{vec}^\top(X)(B\boxtimes B)\mathrm{vec}(X)$$
The box-product essentially behaves like the Kronecker product and satisfies the following properties:
\begin{eqnarray}
A\boxtimes(B\boxtimes F) &=& (A\boxtimes B)\boxtimes F \\
(A\boxtimes B)( C\boxtimes D) &=& (A D)\otimes(BC)\\
( A\boxtimes B)^\top &=& B^\top\boxtimes A^\top\\
(A\boxtimes B)^{-1} &=& B^{-1}\boxtimes A^{-1}\\
\mathrm{trace}(A\boxtimes B) &=& \mathrm{trace}(AB)\\
(A\boxtimes B)\mathrm{vec}(X) &=& \mathrm{vec}(BX^\top A^\top).
\end{eqnarray}
In addition to these Kronecker and box products can easily be multiplied using the following rules:
\begin{eqnarray}
(A\boxtimes B)(C\otimes D) &=& (AD)\boxtimes(BC)\\
(A\otimes B)(D\boxtimes C) &=& (AD)\boxtimes(BC)\\
(A\boxtimes B)(C\otimes D) &=& (A\otimes B)(D\boxtimes C)\\
(A\boxtimes B)(C\boxtimes D) &=& (A\otimes B)(D\otimes C)
\end{eqnarray}
For two two-by-two matrices the box product can explicitly be written 
$$
A\boxtimes B = \begin{pmatrix} 
a_{11}b_{11} & a_{12}b_{11} & a_{11}b_{12} & a_{12}b_{12} \\
a_{11}b_{21} & a_{12}b_{21} & a_{11}b_{22} & a_{12}b_{22} \\
a_{21}b_{11} & a_{22}b_{11} & a_{21}b_{12} & a_{22}b_{12} \\
a_{21}b_{21} & a_{22}b_{21} & a_{21}b_{22} & a_{22}b_{22} \\
\end{pmatrix}.
$$
and here's an example stride-permutation
$$
L_2^6=I_2\boxtimes I_3 = \left(
\begin{array}{llllll}
 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1
\end{array}
\right).
$$
Finally, I should mention that there is actually a mistake in the answer.  My convention has been to use row-wise concatenation of matrices and the definition above, from which easily follows:
\begin{eqnarray*}
\mathrm{trace}(BXBX) &=& \mathrm{vec}^\top((BXB)^\top)\mathrm{vec}(X)\\
&=& \mathrm{vec}^\top(B^\top X^\top B^\top)\mathrm{vec}(X)\\
&=& \mathrm{vec}^\top(X) B\boxtimes B^\top \mathrm{vec}(X)
\end{eqnarray*}
This answer makes more sense, as the matrix $B\boxtimes B^\top$ is always symmetric, whereas $B\boxtimes B$ is not.
I struggled with questions similar to the one you just made, and the box product has been a great help to me.  I hope this was of help to you.
A: Just think of the matrix $U = XB.$ Your matrix is $U^2,$ so write down the expression for the trace of the square, then substitute, then find the coefficient of $x_{ij}.$ It's tedious, but not difficult.
