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 $$\mathbf{b}'\mathbf{\Delta}\mathbf{C}'(\mathbf{C}\mathbf{\Delta}\mathbf{C}')^{-1}=\mathbf{a}'\mathbf{D}\mathbf{C}'(\mathbf{C}\mathbf{D}\mathbf{C}')^{-1}.$$

• Would you mind sharing a little about where is this conjecture coming from? Also, are you looking for all solutions or just some of them.
– KBS
Feb 26, 2022 at 11:47
• @KBS It comes from trying to establish the equivalence between two optimal forecasting problems with Gaussian shocks. Each side of the equation is the optimal forecast equation. $\mathbf{D}$ and $\mathbf{\Delta}$ are covariance matrices for uncorrelated shocks (thus diagonal and positive definite).
– mzp
Feb 26, 2022 at 12:21
• No, I meant the condition of the conjecture. Where does the condition $a^Tb>0$ come from?
– KBS
Feb 26, 2022 at 12:30
• One solution would be enough by the way. The condition $\mathbf{a}'\mathbf{b}>0$ is just associated with the setup we are interested in, so I probably should just have put it together with the other definitions. My guess is actually that it would be possible to find $\mathbf{\Delta}$ even if it does not hold.
– mzp
Feb 26, 2022 at 12:44

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.