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Suppose $A, B$ are positive definite Hermitian matrices, $U$ is a unitary matrix such that $AUB$ is Hermitian. The spectral radius of a square matrix $X$ is denoted by $\rho(X)$. In my study, I guessed an inequality $$\rho(U^*AU+B)\le \rho(A+B).$$ That is, the largest eigenvalue of $U^*AU+B$ is no larger than the largest eigenvalue of $A+B$. How to prove it?

The condition $AUB$ being Hermitian is indispensible, but I have no clue how to use it.

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    $\begingroup$ Denoting $C=U^*AU$ we get $UCB=AUB=R$ is Hermitian, so $CB=U^*R$. That is, $U$ comes from the polar decomposition of $CB$, if $R$ is additionally assumed to be positive definite. $\endgroup$ – Fedor Petrov Apr 9 '16 at 9:32
  • $\begingroup$ Another observation (though somewhat beside the point) is that $U(n)$ has (real) dimension $n^2$, and you're imposing exactly that many conditions, so typically there will probably only be finitely many $U$'s satisfying your assumption for given $A,B$. $\endgroup$ – Christian Remling Apr 9 '16 at 18:19
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    $\begingroup$ Can you say what your reasons are to believe that the inequality is true? $\endgroup$ – Iosif Pinelis Apr 11 '16 at 4:09
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    $\begingroup$ Some easy suggestions: (i) note that $U=BC(CB^2C)^{-1/2}$, where $C$ is as in the comments by Fedor Petrov and M. Lin. (ii) Let $L:=C+B$ and $R:=UCU^*+B$. Then it suffices to show that $\text{tr}(L^k)\le\text{tr}(R^k)$ for all (large enough) natural $k$. This inequality trivially holds for $k=1$, and numerical experiments suggests it holds for all natural $k$. An advantage of the latter inequality is that it is polynomial in the elements of the matrices $L$ and $R$. $\endgroup$ – Iosif Pinelis Apr 11 '16 at 18:47
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    $\begingroup$ I am generating $U$ in the manner suggested by the discussion above. We want $$AUB=BU^*A$$ so assume that $AU=BH$ where $H$ is a Hermitian matrix. Then $A^2=BH^2B$, and I get $H$ from taking the positive square root of $B^{-1}A^2B^{-1}$, and then $U=A^{-1}BH$. As the discussion above suggests, different roots could be used for generating $H$. And having thought about this a little more, I would say the resulting iterations seems more like a modified Duggal iteration. $\endgroup$ – rucarden Apr 26 '16 at 14:09
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I still don't have enough reputations points to comment but here is another related iteration. Geometrically the inequality says that using the matrix $U$ to transform $A$ moves the hyperellipse associated with $A$ further away from the hyperellipse associated with $B$. This has something to do with the angles between the eigenvectors of $A$, $U^*AU$ and $B$.

If we diagonalize $A=V_A\Lambda_AV_A^*$ and $B=V_B\Lambda_BV_B^*,$ then $$AUB=V_A\Lambda_A V_A^*UV_B\Lambda_B V_B^*$$. Taking a similarity transformation with $V_A\Lambda_A$ and letting $\Theta_2=V_A^*UV_B$ and $\Theta_1=V_A^*V_B$ denote the matrices whose entries have the information regarding the angles between the eigenvectors of $A$ , $U^*AU$ and $B$, then we have $$H_1=\Theta_2\Lambda_B \Theta_1^*\Lambda_A^{-1} $$ which is a Hermitian matrix. We have $$ H_1\Lambda_A\Theta_1=\Theta_2\Lambda_B$$ At the next iteration we would have $$ H_2\Lambda_A\Theta_2=\Theta_3\Lambda_B$$ Eliminating $\Theta_3$, we have $$ H_2^2=\Lambda_A^{-1}\Theta_2\Lambda_B^2\Theta_2^*\Lambda_A^{-1},$$ and now eliminating $\Lambda_B$ and $\Theta_2$ and taking a square root, $$H_2=(\Lambda_A^{-1}H_1\Lambda_A^2H_1\Lambda_A^{-1})^{1/2}.$$ $\Lambda_B$ is absent from this iteration though it can be recovered. It seems that the 2-norm of the $H_i$ is increasing and that $H_i$ must converge to $\Lambda_B\Lambda_A^{-1}$ where the eigenvalues of $A$ are in ascending order and the eigenvalues of $B$ are in descending order.

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  • $\begingroup$ Thanks for the comments... it is not clear to me what this would lead to? $\endgroup$ – M. Lin May 6 '16 at 2:12

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