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Suppose that we have symmetric positive definite matrices $A,B,D \in R^{n \times n}$. Further, we have $$\begin{bmatrix}A&C\\C^T &B \end{bmatrix} > 0$$ and also $$\begin{bmatrix}A B^{-1} A &C\\C^T & B A^{-1} B \end{bmatrix} > 0$$ holds.

Does anyone knows whether $$\begin{bmatrix}(A+D) (B+D)^{-1} (A+D) & C+D\\C^T +D & (B+D) (A+D)^{-1} (B+D) \end{bmatrix} > 0$$ holds? Thank you!

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  • $\begingroup$ Have you done any computer experiment to see that if this conjecture is convincing or not? $\endgroup$
    – user1551
    Jan 13, 2015 at 20:27
  • $\begingroup$ I did some tests and it seems to work, but I do not know how to show this. $\endgroup$
    – Ben
    Jan 14, 2015 at 14:15

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The conjecture is wrong. It should be easy to generate a random counterexample using a computer. E.g. I obtained the followings: $$ \begin{bmatrix}A&C\\ C^T&B\end{bmatrix}= \left[\begin{array}{cc|cc} 7&1&0&0\\ 1&1&0&2\\ \hline 0&0&3&1\\ 0&2&1&7 \end{array}\right], \ D=\begin{bmatrix}5&-3\\ -3&3\end{bmatrix}. $$ you may verify that $A,B,D$ as well as the first two block matrices in your question are positive definite, but the last block matrix in your question has a negative eigenvalue.

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  • $\begingroup$ Thanks! I made a mistake in the experiments. $\endgroup$
    – Ben
    Jan 14, 2015 at 16:39

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