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Let $\mathbf{Z,R}$ two Hermitian semidefinite positive matrices with all eigenvalues larger than one. Intuition drives me that

$\mathbf{R}^{-1/2}\mathbf{Z} \left(\mathbf{R}^{-1/2}\right)^H - \mathbf{Z} \preceq \mathbf{0}$

Any idea of how to proceed with the inequality verification?

Thank you.

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2 Answers 2

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Let $A$ and $B$ be Hermitian positive definite matrices. Then, the following matrix \begin{equation*} \begin{pmatrix} A & X\\ X^* & B \end{pmatrix} \end{equation*} is positive definite if and only if $X=A^{1/2}ZB^{1/2}$ for some $Z$ that satisfies $\|Z\| \le 1$ (thus, the range of $X$ is a subspace of the range of $A$, and the range of $X^*$ is a subspace of the range of $B$).

To this matrix, apply the Schur-complement's lemma to conclude that \begin{equation*} A \ge X^* B^{-1}X. \end{equation*}

Using $X=R^{-1/2}$ and $B=A^{-1}$, we must therefore have $R^{-1/2} = A^{1/2}ZA^{-1/2}$ for some contraction $Z$. If $R$ satisfies such an equality, the original claim will be true.

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  • $\begingroup$ Nice derivation. Can I include you in the paper's acknowledgement ? $\endgroup$
    – mikitov
    Mar 20, 2015 at 7:57
  • $\begingroup$ @mikitov: this is somewhat standard in matrix analysis, so I would at best say: "We thank Mathoverflow (linking to this post) for bringing to our attention a standard result that helps...." $\endgroup$
    – Suvrit
    Mar 20, 2015 at 13:03
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Not true. Try $$ R = \pmatrix{4 & 0\cr 0 & 10^4\cr},\ Z = \pmatrix{110 & 100\cr 100 & 100\cr} $$ $$ R^{-1/2} Z R^{-1/2} - Z = \pmatrix{-82.5 & -99.5 \cr -99.5 & -99.99\cr}$$ which is indefinite (its determinant is negative).

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  • $\begingroup$ Right, but I guess the ordering defined with arbitrary matrices is preserved. $\endgroup$
    – mikitov
    Mar 20, 2015 at 7:59

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