# Questions tagged [block-matrices]

For questions about matrices which are defined block wise, like $\pmatrix{A&B\\ C&D}$ where $A,B,C$ and $D$ are themselves matrices. Use this tag with (matrices), and often with (linear-algebra).

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### Interesting facts about the matrix $A^\ast = \left(\begin{smallmatrix} A & -aI \\ a I & A \end{smallmatrix}\right)$ where $A$ is negative definite

I have a fairly general question about facts of the following matrix $$A^\ast = \begin{pmatrix} A & -aI \\ a I & A \end{pmatrix}$$ where $A \in \mathbb{R}^{n \times n}$ is negative ...
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### What can we say for two orthogonal matrices to commute?

Suppose that $Q$ is a block diagonal matrix $\mathrm{diag}(Q_1,\dots,Q_r)$ where $Q_i$ is an orthogonal matrix for $i=1,\dots,r$. Let $V$ be an orthogonal matrix such that $V^TQV=Q$. Can we say ...
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### A counter-example to disprove that not every adjacency matrix can be made block-diagonal?

I have a bunch of clusters of proteins (a disconnected graph) and wanted to present this data as an adjacency matrix for various reasons and have been looking for a way to make these adjacency ...
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### How to take the determinant of a partitioned matrix with vectors.

I am attempting to complete the exercises in my textbook on matrix differential calculus. This question is giving me some trouble. These are the problems I am attempting to complete: If $|A| \ne 0$...
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### $\operatorname{rank}(M)=\operatorname{rank}(A)+\operatorname{rank}(B-A)$

Let $A, B \in M_n,$ and $M= \begin{pmatrix} A & A \\ A& B \end{pmatrix}\\$ I have to prove that $$\operatorname{rank}(M)=\operatorname{rank}(A)+\operatorname{rank}(B-A)$$ Any ideas?
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### Looking for the name of block diagonal decomposition

I remember there exist a decomposition such that it transforms any given matrix to a block diagonal matrix, and each block having single eigenvalue. I couldn't find the name, please help me. I tried ...
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### Inverse of a peculiar block matrix

I'd like to find the inverse of a certain 2x2 block matrix. Since its structure is peculiar, the usual inverse formula cannot be applied. However, for the same reason, I think there is a way to ...
Let $a, b$ and $c$ be three distinct positive integers. Consider the following $2k\times 2k$ matrix: \begin{equation} A=\begin{bmatrix} a&b&\cdots&b&c&c&\cdots&c\\ b&a&...