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.


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

  • $\begingroup$ 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. $\endgroup$
    – KBS
    Feb 26, 2022 at 11:47
  • $\begingroup$ @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). $\endgroup$
    – mzp
    Feb 26, 2022 at 12:21
  • $\begingroup$ No, I meant the condition of the conjecture. Where does the condition $a^Tb>0$ come from? $\endgroup$
    – KBS
    Feb 26, 2022 at 12:30
  • $\begingroup$ 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. $\endgroup$
    – mzp
    Feb 26, 2022 at 12:44

1 Answer 1


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.


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