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.

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 9, 2016 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 $A,B$. $\endgroup$ Apr 9, 2016 at 18:19
  • 1
    $\begingroup$ Can you say what your reasons are to believe that the inequality is true? $\endgroup$ Apr 11, 2016 at 4:09
  • 2
    $\begingroup$ 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 $R$. $\endgroup$ Apr 11, 2016 at 18:47
  • 1
    $\begingroup$ 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, 2016 at 14:09

1 Answer 1


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.

  • $\begingroup$ Thanks for the comments... it is not clear to me what this would lead to? $\endgroup$
    – M. Lin
    May 6, 2016 at 2:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.