Some help for a non-expert, please?
Let $\mathbf{b}= [b_1\cdots b_m \cdots b_M]^T$ and $\mathbf{c}= [c_1\cdots c_m \cdots c_M]^T$.
Given the identity
$\mathbf{D}^T \, \mathrm{diag}(\mathbf{b}) \, \mathbf{D} = \sum_{m=1}^{M}b_mc_m\mathbf{A}_m,$
where $\mathbf{A}_m \in \Re^{N\times N}, \mathbf{A}_m \geq 0$.
Is there a way to solve for $\mathbf{D}$ in which we do not need to know $\mathbf{b}$? In other words: my intuition tells me that there could be a closed-form expression for $\mathbf{D}$ in which $\mathbf{b}$ does not appear.
Thanks in advance,
I will self-answer my question, but I would thank anyone who can provide a proof, even if the proof relies on basic linear algebra... So:
We write
$\mathbf{D}^T \, \mathrm{diag}(\mathbf{b}) \, \mathbf{D} = \sum_{m=1}^{M} (\mathbf{D}_{m*})^T\mathbf{D}_{m*}b_m$,
so therefore, since
$\sum_{m=1}^{M} (\mathbf{D}_{m*})^T\mathbf{D}_{m*}b_m = \sum_{m=1}^{M}b_mc_m\mathbf{A}_m$,
then
$(\mathbf{D}_{m*})^T\mathbf{D}_{m*} = c_m\mathbf{A}_m$.
This means that we can construct matrix $\mathbf{D}$, element-wise, as
$\mathbf{D}_{mn} = (c_m \mathbf{A}_{m,nn})^{1/2}$,
where $\mathbf{A}_{m,nn}$ represents the $n$-th diagonal entry of matrix $\mathbf{A}_m$.