Solving for a Diagonal Positive Definite Matrix: $\Delta$ such that $b'\Delta C'(C\Delta C')^{-1}=a'DC'(CDC')^{-1}$ Any help with the following conjecture would be highly appreciated. I have a hard time figuring out how to start. Though counterexamples would be certainly helpful, if the conjecture is not always true I would be most interested in the weakest set of assumptions under which it is true.

Let

*

*$\mathbf{a}$ and $\mathbf{b}$ be $m\times 1$ vectors,

*$\mathbf{C}$ be an arbitrary $n\times m$ matrix of rank $n$, with $n\le m$,

*and $\mathbf{D}$ be an arbitrary diagonal positive definite $m\times m$ matrix.

Conjecture:$\quad$If $\mathbf{a}'\mathbf{b}>0$, then there exists a diagonal positive definite $m\times m$ matrix $\mathbf{\Delta}$ such that
$$\mathbf{b}'\mathbf{\Delta}\mathbf{C}'(\mathbf{C}\mathbf{\Delta}\mathbf{C}')^{-1}=\mathbf{a}'\mathbf{D}\mathbf{C}'(\mathbf{C}\mathbf{D}\mathbf{C}')^{-1}.$$
 A: I have a beginning of an answer. I will update it when I make some progress.
Assume $\eta\in\mathbb{R}^m$ such that $C\eta=0$. Then, one can define $a-b=D^{-1}\eta$ and $\Delta=\alpha D$ for any $\alpha>0$. Note that there is no reason that $a^Tb>0$ be satisfied in this case.
It is also possible to use Kronecker calculus for that problem. Define $M:=DC^T(CDC^T)^{-1}C$. Then, the equality writes
$$(b^T-a^TM)\Delta C^T=0.$$
We can immediately notice that if $b^T-a^TM=0$, then the equality is satisfied for all diagonal $\Delta$'s with positive diagonal entries. Note that we can always pick $a$ and $b$ such that this holds true.
Assuming this solution is not satisfactory, we can rewrite the equality as
$$(C\otimes (b^T-a^TM))\mathrm{vec}(\Delta)=0$$
where $\mathrm{vec}$ denotes the vectorization operator. In order to capture the diagonal structure of $\Delta$, we introduce the matrix $J$ such that $\mathrm{vec}(\Delta)=Jx$ where $x\in\mathbb{R}^m$ is a positive vector that contains the diagonal entries of $\Delta$.
So, this leads to the problem of finding such an $x$ such that
$$(C\otimes (b^T-a^TM))Jx=0.$$
So, the problem becomes, in the end, finding a positive vector in the null-space of some matrix. However, it does not seem possible to conclude on anything general based on this formulation. However, this can be numerically checked quite easily.
