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Let $M=\left[\begin{array}{cc}A & B \\B^T & C \end{array}\right]\in\mathbb{R}^{(n+m)\times (n+m)}$ be a symmetric positive definite matrix, where $n\geq m$, $A\in\mathbb{R}^{n\times n}$, $C\in\mathbb{R}^{m\times m}$ are symmetric positive definite matrices and $B\in\mathbb{R}^{n\times m}$ be a full rank matrix, i.e., $\text{rank}(B)=m$. What is the spectrum of $M$ in terms of its blocks? I am particularly interested in having a lower bound on the minimum eigenvalue of $M$. I would highly appreciate any suggestions or references.

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$X:=\left[\begin{array}{cc}A & \mathbf 0 \\\mathbf 0 & C \end{array}\right]=M-\left[\begin{array}{cc}\mathbf 0 & B\\ B^T &\mathbf 0 & \end{array}\right]$

$\lambda_{min}\Big(M\Big)$
$= \lambda_{min} \left( X + \left[\begin{array}{cc}\mathbf 0 & B\\ B^T &\mathbf 0 & \end{array}\right]\right)$
$\geq \lambda_{min} \left( X\right) + \lambda_{min} \left(\left[\begin{array}{cc}\mathbf 0 & B\\ B^T &\mathbf 0 & \end{array}\right]\right)$
$= \lambda_{min} \left( X\right) + -\lambda_{max} \left(BB^T\right)^\frac{1}{2}$
$= \lambda_{min} \left( X\right) -\sigma_{max} \left(B\right)$
$= \min\Big(\lambda_{min}( A),\lambda_{min}(C)\Big) -\sigma_{max} \left(B\right)$
This bound is sharp --- if you play around with projection matrices, it's easy to construct cases where this is met with equality.

Something else of interest
$M \succeq X$ i.e. $M$ majorizes $X$ and

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  • $\begingroup$ Hi there, the matrix M, in this case, is positive definite. However, the bound given by your solution gives me a negative one. I wonder in what case M majorizes X? $\endgroup$
    – math1223
    Apr 28, 2020 at 22:41
  • $\begingroup$ your original post has $B$ as an arbitrary full rank matrix -- there's not much more that can be said with the problem statement. $M$ always majorizes $X$. $\endgroup$ Apr 29, 2020 at 4:54
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    $\begingroup$ @gen The standard reference is going to be Marshall & Olkin. But you should be able to prove this on your own in at most 3 lines. First check that their traces are equal. Next consider the set $S_r$ of rank $r$ orthogonal projections that are block diagonal --in particular with same block structure as $X$. For well chosen $P\in S_r$ we have $\sum_{k=1}^r \lambda_k^{(X)}=\text{trace}\big(PX\big) =\text{trace}\big(PM\big)\leq \sum_{k=1}^r \lambda_k^{(M)}$ by von Neumann Trace Inequality. (note: eigenvalues are always labelled from largest to smallest). $\endgroup$ Oct 5, 2022 at 17:35
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    $\begingroup$ the specifics are different but conceptually it is the exact same idea as here: math.stackexchange.com/questions/2700488/… ... i.e. adding zero does not change the trace. $\endgroup$ Oct 5, 2022 at 17:39
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    $\begingroup$ @DuttaA you stated "your claim 𝑀⪰𝑋 implies $\lambda_{\min} (M) \geq \lambda_{\min} (X)$". This is false. $\succeq$ Can mean many things. When I wrote this I indicated that $\succeq $ denotes majorization. (More technically the eigenvalues of $M$ strongly majorize the eigenvalues of $X$, which is a cleaner statement than what I originally wrote). $\endgroup$ Nov 5, 2022 at 19:34

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