# Spectrum of SPD block matrix

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.

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

$$\lambda_{min}\Big(M\Big)$$
$$= \lambda_{min} \left( X + \left[\begin{array}{cc}\mathbf 0 & B\\ B^T &\mathbf 0 & \end{array}\right]\right)$$
$$\geq \lambda_{min} \left( X\right) + \lambda_{min} \left(\left[\begin{array}{cc}\mathbf 0 & B\\ B^T &\mathbf 0 & \end{array}\right]\right)$$
$$= \lambda_{min} \left( X\right) + -\lambda_{max} \left(BB^T\right)^\frac{1}{2}$$
$$= \lambda_{min} \left( X\right) -\sigma_{max} \left(B\right)$$
$$= \min\Big(\lambda_{min}( A),\lambda_{min}(C)\Big) -\sigma_{max} \left(B\right)$$
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

• 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? Apr 28, 2020 at 22:41
• 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 $X$. Apr 29, 2020 at 4:54
• @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). Oct 5, 2022 at 17:35
• 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. Oct 5, 2022 at 17:39
• @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). Nov 5, 2022 at 19:34