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Let $A$ and $B$ be $n\times n$ Hermitian positive definite matrices and $k>0$ real. Then $A^k$ is well-defined and experimentally, we have $$\det(A^k+BABA^{-1})\geqslant \det(A^k+BA^{-1}BA),$$or equivalently $$\det(A^k+A^{-1}BAB)\geqslant \det(A^k+ABA^{-1}B).$$ Note that this would imply the inequality $\color{green}{\det (A^2+BAAB)\geqslant \det(A^2+ABBA)}$ found (but presumably unproved) by M. Lin, which is mentioned in a comment here.

One striking fact is that $B$ (or $A$, same outcome) can be multiplied by a positive factor, which means that the inequality remains valid for any "proportionality" between the two terms of each sum. Indeed, for any $\lambda>0$, we have equivalently $$\det(A^k+\color{red}\lambda BABA^{-1})\geqslant \det(A^k+\color{red}\lambda BA^{-1}BA).$$

Any ideas how to prove this conjecture?

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  • $\begingroup$ I have some other generalizations based on your very interesting determinantal conjectures --- I wish I had time to think about these. Do you have numerical evidence for these? The second one suggests that perhaps there is even a Löwner dominance somewhere that yields the ineq for all $\lambda\ge 0$. $\endgroup$ – Suvrit Jun 23 '17 at 11:06
  • $\begingroup$ @Suvrit Yes I wouldn't have posted that without checking first for several thousands of random matrices of different sizes and several big and small values of $k$. I almost thought my few lines of PARI code had a bug... :-) $\endgroup$ – Wolfgang Jun 23 '17 at 11:14
  • $\begingroup$ have you tried Tao's strategy: proving that the spectrum of $C=A^{-(k+1)/2}BABA^{-(k+1)/2}$ is majorated by the spectrum of $D=A^{-(k-1)/2}BA^{-1}BA^{-(k-1)/2}$? $\endgroup$ – Fedor Petrov Jun 23 '17 at 11:45
  • $\begingroup$ @FedorPetrov That looks like an appealing idea, but I already have a hard time following how he re-arranges the traces. And no idea how to use that method with inverses, let alone arbitrary powers, even the k=2 case. $\endgroup$ – Wolfgang Jun 23 '17 at 12:12
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    $\begingroup$ Sorry to have confused you, I had known this $\det (A^2+BAAB)\geqslant \det(A^2+ABBA)$. It follows from COROLLARY 2.3 of the paper M. Lin, H. Wolkowicz, An eigenvalue majorization inequality for positive semidefinite block matrices, Linear Multilinear Algebra 60 (2012), no. 11-12, 1365–1368. tandfonline.com/doi/abs/10.1080/03081087.2011.651723 $\endgroup$ – M. Lin Jun 25 '17 at 0:00
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Let $C = A^{-(k+1)/2} B A B A^{-(k+1)/2}$ and $D = A^{-(k-1)/2} B A^{-1} B A^{-(k-1)/2}$ .

We have to show that $det(I+C) \ge det(I+D)$ .

Now my goal is to apply equation (5.21) in Ando, Majorizations and Inequalities in Matrix Theory, http://ac.els-cdn.com/0024379594903417/1-s2.0-0024379594903417-main.pdf?_tid=1af72ca2-58c4-11e7-a518-00000aab0f01&acdnat=1498298674_bf9275307496eb46505ba4b5b84a7dcc :

Choose $E = A^{-(k+1)/2} B A^{-(k+1)/2}$ and $F = A^{k/2+1}$.

Then $C^{1/2} = \vert F E \vert$ and $D^{1/2} = \vert F^{a_1} E^{b_1} F^{a_2} E^{b_2} \vert$ , where $a_1 = (k/2)/(k/2+1)$, $b_1 = 1$, $a_2 = 1/(k/2+1)$, $b_2 = 0$.

Since $0 \le a_1 \le b_1$ and $0 \le a_2$ and $0 \le b_2$ and $a_1 + a_2 = b_1 + b_2 = 1$ it follows that $D^{1/2} \prec_{\log} C^{1/2}$ and therefore $D \prec_{\log} C$.

Since $log(1+e^x)$ is convex in $x$ it follows that $I+C$ weakly log majorizes $I+D$ and therefore $det(I+C) \ge det(I+D)$ .

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  • $\begingroup$ Shouldn't it be (k+1)/2 in the definition of E? $\endgroup$ – Wolfgang Jun 24 '17 at 14:40
  • $\begingroup$ @Wolfgang : Yes, corrected. $\endgroup$ – jjcale Jun 24 '17 at 17:00
  • $\begingroup$ @jjcale, would you try Conjecture 4.2 in the paper worldscientific.com/doi/abs/10.1142/S0219199716500449 $\endgroup$ – M. Lin Jun 25 '17 at 0:09
  • $\begingroup$ @M.Lin, can you state the conjecture here, please? $\endgroup$ – Wolfgang Jun 25 '17 at 17:26
  • $\begingroup$ Here is the link to the arXiv preprint arxiv.org/pdf/1604.04141v1.pdf $\endgroup$ – M. Lin Jun 26 '17 at 5:02
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EDIT (added some clarifications). The argument below provides a self-contained proof.

Introduce the shorthand $C^{-2}=A^{k+1}$. We need to show that \begin{equation*} \det(I+ CBABC) \ge \det(I + CABA^{-1}BAC). \end{equation*}

We will prove this inequality by establishing the log-majorization \begin{equation*} \prod\nolimits_{j=1}^k\lambda_j(CABA^{-1}BAC) \le \prod\nolimits_{j=1}^k\lambda_j(CBABC),\quad 1\le j \le n. \end{equation*} This log-majorization is equivalent to $\lambda_1(\wedge^k (CABA^{-1}BAC)) \le \lambda_1(\wedge^k(CBABC))$ (for $1\le k\le n$). It suffices to show this for $k=1$; the general case follows similarly upon exploiting the multiplicativity of the wedge product.

Thus, to prove $\lambda_1(CABA^{-1}BAC) \le \lambda_1(CBABC)$, we use its scale independence and observe that for positive matrices $X$ and $Y$, we have \begin{equation*} [Y \le I \implies X \le I] \implies \lambda_1(X)\le \lambda_1(Y). \end{equation*}

Thus, to prove the $\lambda_1$ inequality, it suffices to prove that \begin{equation*} CBABC \le I \Leftrightarrow\ \begin{bmatrix} A^{k+1} & B\\ B & A^{-1} \end{bmatrix} \ge 0\quad\implies CABA^{-1}BAC \le I. \end{equation*} But $\begin{bmatrix} A^{k+1} & B\\ B & A^{-1} \end{bmatrix} \ge 0$ only if $B\le \sqrt{A^{k+1}A^{-1}}=A^{k/2}$; and if this is so, then it must also be the case that $\begin{bmatrix} A^{k-1} & B\\ B & A \end{bmatrix} \ge 0$. Notice now that $A^{k-1} = C^{-2}A^{-2}$, and apply Schur complements to the latter matrix inequality to obtain \begin{equation*} BA^{-1}B \le C^{-2}A^{-2}\Leftrightarrow CABA^{-1}BAC \le I. \end{equation*} Thus, we have shown that $\lambda_1(CABA^{-1}BAC) \le \lambda_1(CBABC)$. In a similar manner we can prove the general case for $k>1$, which ends up establishing the desired log-majorization.

This finishes the proof of the claim because $\lambda(X) \prec_{\log} \lambda(Y) \implies \det(I+X) \le \det(I+Y)$.

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    $\begingroup$ How can you see that $B \le A^{k/2}$ ? $B$ is positive definite but otherwise arbitrary . $\endgroup$ – jjcale Jun 24 '17 at 6:33
  • $\begingroup$ Yes, how can you apply your first lemma if $Y\le I$ doesn't hold? $\endgroup$ – Wolfgang Jun 24 '17 at 7:09
  • $\begingroup$ @jjcale Because it's a conditional statement; if we have $CBABC\le I$, then we have $B\le A^{k/2}$. What we really wanted to prove was the eigenvalue domination; we reduced that to showing that $Y\le I \implies X \le I$ must hold (note that $\lambda(\cdot)$ is positively homogeneous). $\endgroup$ – Suvrit Jun 24 '17 at 12:10
  • $\begingroup$ Actually, I should be a bit more precise, and just write that what I am using is the observation: $(0 \le Y \le I \implies 0 \le X \le I) \implies \lambda_1(X)\le \lambda_1(Y)$ $\endgroup$ – Suvrit Jun 24 '17 at 12:14
  • $\begingroup$ So you mean your proof only works if CBABC≤I? Well, granted, but then that's a very special case only. $\endgroup$ – Wolfgang Jun 24 '17 at 12:55

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