Let $\{C_i\}_{i=1}^N$ be a set of $n\times m$ real matrices of full column-rank and such that $\mathrm{Range}[C_1,C_2,\dots,C_N]=\mathbb{R}^n$, $\{P_i\}_{i=1}^N$ a set of $m\times m$ positive definite real matrices. Moreover, let $Q>0$ be a positive definite $n\times n$ real matrix such that $Q\in\mathrm{Range}\,\mathcal{C}$, where $\mathcal{C}$ is the linear operator $$ \mathcal{C}\colon A\mapsto \sum_{i=1}^N C_i A C_i^\top. $$

Consider the following equation in the unknown variable $X\in\mathbb{R}^{n\times n}$ $$\tag{$\star$}\label{a} \sum_{i=1}^N C_i(C_i^\top X C_i)^{-\frac{1}{2}}P_i(C_i^\top X C_i)^{-\frac{1}{2}}C_i^\top=Q. $$ (Here, $A^{\frac{1}{2}}$ denotes the principal matrix square root of $A\ge 0$.)

My question: Does \eqref{a} always admit a positive definite solution $\bar{X}>0$?

Notice that for the particular case $P_i=I_m$, $i=1,2,\dots,N$, \eqref{a} reduces to $$\tag{$\star\star$}\label{aa} \sum_{i=1}^N C_i(C_i^\top X C_i)^{-1}C_i^\top=Q. $$ In this case, I have some evidence to believe that the answer is in the affirmative. However, the general case is still obscure to me.

EDIT. As Noah Stein correctly noticed in his answer below, the answer is negative for a general $Q>0$. However, I figured out that I forgot a fundamental assumption in my question, which now I added.


1 Answer 1


Let $N=n$, $m=1$, $P_i = 1$ for all $i$, and let the $C_i$ be the standard unit vectors. Then the left hand side of $(\star\star)$ is the matrix whose diagonal entries are the inverses of the diagonal entries of $X$ and whose off-diagonal entries are zero. So there is no solution, positive definite or otherwise, for $Q$ with zeros off the diagonal.

  • $\begingroup$ You are absolutely right! I forgot an assumption in my question. Please see my edit. $\endgroup$
    – Ludwig
    Jan 18, 2017 at 7:39

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