Let $\{V\_i\}\_{i=1}^N$ be a set of $n\times m$, $n\geq m$, real matrices of full column rank and let $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive definite trace-one matrix. Moreover, let $A^{1/2}=(A^{1/2})^\top$ denote the (unique) symmetric square root of a positive semi-definite matrix $A$, $\mathrm{tr}\, A$ the trace of $A$, and $\log A$ the matrix logarithm of $A$. 



> 
> **Question**: Does the inequality
>  $$\tag{$1$} \label{1}
> \sum\_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum\_{j=1}^N V\_i^\top X^{1/2}V\_j(V\_j^\top X V\_j)^{-1} V\_j^\top X^{1/2}V\_i\right) - \sum\_{i=1}^N \mathrm{tr} \log \left(V\_i^\top X V\_i\right) \ge 0
> $$
>  hold true?
> 
> 
> 


**My (very little) progress so far.**
Using the fact that $\mathrm{tr}\log Y+\mathrm{tr}\log Z=\mathrm{tr}\log YZ$ and $-\log\, Y=\log Y^{-1}$ for positive definite $Y$, $Z$, we can rewrite \eqref{1} as 
$$\tag{$2$}
\sum\_{i=1}^N \mathrm{tr}\log \left(\frac{1}{mN}\sum\_{j=1}^N V\_i^\top X^{1/2}V\_j(V\_j^\top X V\_j)^{-1} V\_j^\top X^{1/2}V\_i\left(V\_i^\top X V\_i\right)^{-1}\right) \ge 0.
$$
Also, exploiting the formula $\mathrm{tr}\log Y =\log \det Y$, we get
$$\tag{$3$}
\sum\_{i=1}^N \log\det \left(\frac{1}{mN}\sum\_{j=1}^N V\_i^\top X^{1/2}V\_j(V\_j^\top X V\_j)^{-1} V\_j^\top X^{1/2}V\_i\left(V\_i^\top X V\_i\right)^{-1}\right) \ge 0,
$$
or, equivalently,
$$\tag{$3$}
\prod\_{i=1}^N \det \left(\frac{1}{mN}\sum\_{j=1}^N V\_i^\top X^{1/2}V\_j(V\_j^\top X V\_j)^{-1} V\_j^\top X^{1/2}V\_i\left(V\_i^\top X V\_i\right)^{-1}\right) \ge 1.
$$
I'm stuck at this point. I believe that the problem can be further simplified to some well-known determinant inequality. However, at the moment, I don't know how to proceed. 




---


**Note 1.** This question can be seen as a generalization of [this problem](https://mathoverflow.net/questions/233546/a-log-inequality-for-positive-definite-trace-one-matrices). Indeed, for $m=1$, \eqref{1} reduces exactly to the case treated in that question, which has been proved to be true.


**Note 2.** Numerical simulations suggest that \eqref{1} is very likely to be true.


**Edit.** I edited the inequality by replacing the factor $1/N$ in the first logarithm by $1/(mN)$. This is a somewhat "stronger" version which I believe is the "right" generalization of the aforementioned [problem](https://mathoverflow.net/questions/233546/a-log-inequality-for-positive-definite-trace-one-matrices).



A generalized log inequality for positive definite trace-one matrices
Fedor's proof can be generalized :


Let $S = X^{1/2}$, $U\_i = V\_i S^{1/2}$ and $W\_i = U\_i (U\_i^T S U\_i)^{-1/2}$ .
Then $W\_i^T S W\_i = I\_m$ where $I\_m$ is the $m\times m$ identity .


Since $log(x) \ge 1 - 1/x$ it is enough to show that 
$$tr \sum\_{i=1}^N \mu\_i \le 1$$ 
where 
$$\mu\_i = (\sum\_{j=1}^N W\_i^T W\_j W\_j^T W\_i)^{-1}$$ .


Then 
$$tr \sum\_{i=1}^N \mu\_i = tr \sum\_{i=1}^N \mu\_i^2 \mu\_i^{-1}$$
$$= tr [\sum\_{i=1}^N \mu\_i^2 \sum\_{j=1}^N W\_i^T W\_j W\_j^T W\_i]$$
$$\ge tr \sum\_{i=1}^N \sum\_{j=1}^N W\_i \mu\_i W\_i^T W\_j \mu\_j W\_j^T$$
$$= tr [(\sum\_{i=1}^N W\_i \mu\_i W\_i^T)^2]$$
$$\ge [tr (S \sum\_{i=1}^N W\_i \mu\_i W\_i^T)]^2$$
$$= (tr \sum\_{i=1}^N \mu\_i)^2$$ ,


where I have used that
$$tr (\mu\_i^2 W\_i^T W\_j W\_j^T W\_i) + tr (\mu\_j^2 W\_j^T W\_i W\_i^T W\_j) \ge 2 tr (W\_i \mu\_i W\_i^T W\_j \mu\_j W\_j^T)$$ .


This follows from
$$0 \le tr[(W\_i^T W\_j \mu\_j - \mu\_i W\_i^T W\_j) (\mu\_j W\_j^T W\_i - W\_j^T W\_i \mu\_i)]$$ .


