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Let $A, B, X$ be three real, symmetric positive semidefinite matrices and suppose that the inequality $$ X \preceq \sqrt{A} (\sqrt{A} B \sqrt{A} + I)^{-1} \sqrt{A} $$ holds where $\sqrt{\cdot}$ denotes the positive semidefinite square-root of a matrix.

In the case that $A, B $ are both nonsingular, we can write this inequality equivalently as $$ \begin{pmatrix} X & X \\ X & X \end{pmatrix} \preceq \begin{pmatrix} A & 0 \\ 0 & B^{-1}\end{pmatrix}. \qquad \mbox{$(*)$} $$ (See proof below.)

Question: Is there an analogous "simpler" form when $A, B$ are possibly singular (i.e., positive semidefinite but not necessarily definite)?

Proof of $(*)$: Set $M = \left(\begin{smallmatrix} A^{-1} & I \\ B & - I \end{smallmatrix}\right)$; note that it is nonsingular. Hence the inequality above is equivalent to $$ 0 \succeq M^T \begin{pmatrix} X - A & X \\ X & X - B^{-1} \end{pmatrix} M = \begin{pmatrix} A^{-1}XA^{-1} + BXB + A^{-1}XB + BXA^{-1} - (B + A^{-1})& 0 \\ 0& -(A + B^{-1}) \end{pmatrix} $$ Or equivalently, $$ (B+A^{-1})X (B+A^{-1}) \preceq (B + A^{-1}) $$ Which is in turn equivalent to $$ X \preceq (B + A^{-1})^{-1} = \sqrt{A} (\sqrt{A} B \sqrt{A} + I)^{-1}\sqrt{A}. \hspace{5mm}\square $$

Drew Brady
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1 Answers1

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I am using the formula for the inverse here (Inverse of $I+BA$ is $(I+BA)^{-1} = I-B(I+AB)^{-1}A$) $$ (I+CD)^{-1} = I - C(I+DC)^{-1}D \ . $$

We then have the series of equivalence \begin{align*} X \preceq \sqrt{A} (I + \sqrt{A} B \sqrt{A} ) ^{-1} \sqrt{A} \\ X \preceq \sqrt{A} [I - \sqrt{A}B (I+\sqrt{A}\sqrt{A}B)^{-1} \sqrt{A}] \sqrt{A} & \quad \text{using $C=\sqrt{A}B$ and $D=\sqrt{A}$} \\ X \preceq A(I-B(I+AB)^{-1}A) \\ X \preceq A(I+BA)^{-1} & \quad \text{using $C=B$ and $D=A$} \\ (I+BA)^{T}X(I+BA) \preceq A+A^TBA \\ \begin{pmatrix} I \\ A \end{pmatrix}^T \begin{pmatrix} X-A & XB \\ B^TX & B^TXB-B \end{pmatrix} \begin{pmatrix} I \\ A \end{pmatrix} \preceq 0 \\ \end{align*}

Marc Dinh
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