Eigenvalues of an "Half-Kronecker "Product The Problem:
Given a 2 by 2 matrix $C$(the matrix elements of C are given), and two other
2 by 2 matrices $A$ and $B$(the matrix elements of A and B are given).
Now  we can construct a new matrix $D$, which is given by the direct product
of (the first row of $C$) and $A$, the direct product of (the second row of $C$) and $B$, just like this:
$$D = 
\begin{pmatrix}
c_{11} A & c_{12} A \\
c_{21} B & c_{22} B
\end{pmatrix}$$
Four Bolcks.
(Can we have a better way to write this kind of product?)
As we know, if $A=B$, then the eigenvalues of $D$ are products of eigenvalues
of $C$ and eigenvalues of $A$, 4 eigenvalues in all.
Then, can we know the eigenvalues of $D$? What can we say about the
eigenvalues of $D$? Implicit equation are OK.
Is this problem related to Khatri-Rao products? Is there anyone who
considered this problem and solved it?

It seems very doubtful that there could be a simple closed form for the
eigenvalues in general (i.e. simpler than explicitly taking the characteristic
polynomial and solving this quartic polynomial in radicals).
Case in point: take $$a_{{1,1}}=-3,a_{{1,2}}=3,a_{{2,1}}=0,a_{{2,2}}=1,b_{{1,1}}=-3,b_{{1,2
}}=-1,b_{{2,1}}=-2,b_{{2,2}}=-2,c_{{1,1}}=-2,c_{{1,2}}=3,c_{{2,1}}=-3,
c_{{2,2}}=2
$$
so that $$D = \pmatrix{
6&-6&-9&9\cr0&-2&0&3 \cr 9&3&-6&-2\cr 6&6&-4&-4
\cr}$$
Its characteristic polynomial is $t^4+6 t^3-27 t^2+230 t-300$ which is
irreducible over the rationals.  On the other hand, $A$ and $B$ have
integer eigenvalues ($(1,-3)$ and $(-1,-4)$ respectively.
