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Let $A$ be a SPD matrix, let $D$ be the diagonal of $A$, and let $G$ be an arbitrary matrix of the proper size (not necessarily full rank). Do we have the following inequality$$\det(I+G^\top A^{-1}G) \leq \det(I+G^\top D^{-1}G) $$ I feel like this is not true. We can surely prove that $\det(A^{-1})\leq \det(D^{-1})$, but I don't know if this the inequality hold in general.

Also, if this is not true, what condition do we have to impose to $G, A, D$ for this inequality to hold?

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  • $\begingroup$ Your claim that $\det(A^{-1})\le\det(D^{-1})$ is false. Counterexample: when $A=\pmatrix{2&1\\ 1&2}$, we have $\det(A)^{-1}=\frac13>\frac14=\det(D)^{-1}$. In fact, Hadamard's inequality states that $\det A$ is always bounded above by $\det D$. $\endgroup$ – user1551 Jun 10 '18 at 9:25
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    $\begingroup$ AFAIK the most you can say about the spectra of $A$ and $D$ is that the diagonal entries (eigenvalues) of $D$ are included in the interval defined by the smallest and largest eigenvalues of $A$ (you have a lower bound for $\lambda_\max(A)$ and an upper bound for $\lambda_\min(A)$). That does not tell you anything about how their products compare to each other. $\endgroup$ – Algebraic Pavel Jun 10 '18 at 15:32
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The A produced by @user1551, namely $A=\begin{pmatrix} 2 &2\\ 1 & 2\\ \end{pmatrix}$ combined with $G = I$, produces a counterexample:

$\det(I+G^T A^{-1}G) = 8/3 > 9/4 = \det(I+G^T D^{-1}G)$

$\det(I+G^T A^{-1}G) \le \det(I+G^T D^{-1}G)$ is trivially true when $G =$ zero matrix and/or $A$ is a diagonal matrix.

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