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First, I apologize. This is a reposted version because the previous one had factual errors.

This question follows up on a previously raised question.

Let ${\bf A} \in {\Bbb R}^{n \times n}$ be a symmetric positive definite M-matrix whose non-diagonal entries are non-positive. Let ${\bf B} \in {\Bbb R}^{n \times n}$ and ${\bf C} \in {\Bbb R}^{n \times n}$ be two diagonal matrices, whose diagonal entries are non-negative.

$${\bf M} := \begin{bmatrix} \mathbf{A} & \mathbf{0} \newline \mathbf{0} & \mathbf{A} \newline \end{bmatrix} + \begin{bmatrix} -\mathbf{B} & \mathbf{C} \newline \mathbf{C} & \mathbf{B} \newline \end{bmatrix}$$

Can it be proven that the ${\bf smallest}$ eigenvalue of ${\bf M}$ always decreases with the diagonal entries of ${\bf C}$ increase? (Which has been verified numerically)

K416
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    The proofs in both answers to your previous question can be slightly modified to tackle the present case. – user1551 Mar 28 '25 at 14:00
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    in terms of my proof you are asking if $\mathbf 0 \preceq DYD$ that also obeys the component-wise inequalities $0\leq DYD \leq Z\implies \rho(DYD)\leq \rho(Z)$ when $Z=Z^T$? Well Spectral theorem tells us that for some $\mathbf x$ satisfying $\big\Vert \mathbf x \big\Vert_2=1$ and $\mathbf x \geq \mathbf 0$ component-wise (Perron-Frobenius) we have $\rho(DYD)= \mathbf x^T DYD \mathbf x \leq \mathbf x^T Z\mathbf x\leq \rho(Z)$. I'd like to know where this question is coming from. – user8675309 Mar 28 '25 at 16:02
  • @ user8675309 : Thanks for your direction! This is coming from a problem of the power system, where A plays the role of the nodal admittance matrix and B, C are some different influence factors. – K416 Mar 29 '25 at 14:53
  • @ user1551 : Thanks for your direction and I understood. – K416 Mar 29 '25 at 14:54

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