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Let's say I have a block matrix of the form

$$X = \begin{bmatrix} A & B\\ B^T & C\end{bmatrix}$$

where $A$, $C$, and $X$ are all positive definite. I have bounds on both the minimum and maximum eigenvalues of $A$, $C$, and $X$. I am wondering what I can say about the maximum singular value of $B$.

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  • $\begingroup$ Note that $X$ is psd iff $B=A^{1/2}ZC^{1/2}$ for a contraction $Z$ (i.e., $\|Z\|\le 1$). This equality, allows obtaining easy bounds on $\|B\|$. Many other such results can be proved, perhaps you have something more specific in mind? $\endgroup$ – Suvrit Feb 10 '18 at 15:20
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Just a quick lazy answer.

By the interlacing property of Schur complements, for a vector $v$ with unit norm one has $\lambda_{\min}(X) \leq \lambda_{\min}(A-B C^{-1}B^T) \leq v^TAv - v^TB C^{-1} B^T v \leq \lambda_{\max}(A) - \frac1{\lambda_{\max}(C)}\|B^T v\|^2$, which gives the bound $$ (\sigma_{\max}(B))^2 \leq (\lambda_{\max}(A) - \lambda_{\min}(X))\lambda_{\max}(C). $$ I hope I'm not mixing up any min/max here, but in any case you get the idea of the reasoning.

It should be possible to find examples in which all equalities hold. Probably a similar inequality can be set up for the other direction, too ($\sigma_{\max}(B) \geq \sigma_{\min}(B) \geq \dots$).

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    $\begingroup$ It seems that the bound you get should be $\sigma_{\max}(B)^2 \leq (\lambda_\max(A)-\lambda_\min(X))\lambda_\max(C)$? $\endgroup$ – Simon L Rydin Myerson Mar 16 '18 at 16:00
  • $\begingroup$ @SimonLRydinMyerson Yes, that looks correct. Thanks for your fixes; this answer was written very hastily. $\endgroup$ – Federico Poloni Mar 16 '18 at 16:25

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