# Questions tagged [block-matrices]

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### Proof that block matrix has determinant $1$

The following real $2 \times 2$ matrix has determinant $1$: $$\begin{pmatrix} \sqrt{1+a^2} & a \\ a & \sqrt{1+a^2} \end{pmatrix}$$ The natural generalisation of this to a real $2 \times 2$ ...
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### Eigenvalues in unit disk for a 2×2 block matrix

Crossposted from Mathematics. Consider the $2\times 2$ matrix \begin{align*} Q = \begin{bmatrix} 1 & 1 \\ 0 & a \end{bmatrix} - \epsilon \begin{bmatrix} 1 & 0 \\ ...
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### Eigenvalues of a block matrix with zero diagonal blocks

Suppose $A$ is a $k_1\times k_2$ matrix with real entries, $k_1<k_2$. Let $M$ be the matrix \begin{equation} M:=\begin{pmatrix} 0_{k_1} & A\\ A^\top & 0_{k_2} \end{pmatrix}, \end{equation} ...
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### Off-diagonalize a matrix

Consider a self-adjoint matrix $M$ that has block form $$M = \begin{pmatrix} M_{11} & M_{12} \\ M_{12}^* & M_{11} \end{pmatrix}.$$ I am wondering if there exists any criterion to decide if ...
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### Solve linear overdetermined system from subsystems that compose it

This is my first MathOverflow post: I apologize if my message is lacking of something. I also posted this question in Mathematics Stack Exchange, but as I haven't seen an answer I post it here. ...
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### Pseudo-inverse of a column partitioned matrix

Given a $nm \times m$ matrix $A = \begin{pmatrix} A_1 \\ A_2 \\ \vdots \\ A_n\end{pmatrix}$ over $\mathbb{C}$, where $A_i$'s are $m \times m$ and $rank(A) = m$, is there an expression for the pseudo-...
1 vote
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### Determinant diagonal blocks compound matrix [closed]

Good afternoon, I would like to prove the equation \begin{equation} \begin{vmatrix} b_{1,1}I_d & b_{1,2}I_d & \cdots & b_{1,r}I_d \\ b_{2,1}I_d & b_{2,2}I_d & \cdots & b_{2,r}...
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### Algebra of block matrices with scalar diagonals

I am interested in block matrices $A$, that is $A\in M_{n\times n}(R)$ where $R=M_{s\times s}(k)$ and $k$ is a field, such that for every positive integer $m$ the matrix $A^m$ has only scalar blocks ...
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### Is there a formula for the determinant of a block matrix of this kind?

I am looking for an expression that gives the determinant of a matrix of the form \begin{bmatrix} A & B & 0 & \dots & 0 & C \\ B & A & B & & 0 & 0 \\ 0 & ...
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How can we show that there are not defective eigenvalues for this square block matrix of dimension $2d \times 2d$: \begin{bmatrix} A&B\\-B& 0 \end{bmatrix} where A, B are real matrices, $A =\... 7 votes 1 answer 1k views ### Block matrices and their determinants For$n\in\Bbb{N}$, define three matrices$A_n(x,y), B_n$and$M_n$as follows: (a) the$n\times n$tridiagonal matrix$A_n(x,y)$with main diagonal all$y$'s, superdiagonal all$x$'s and subdiagonal ... 4 votes 0 answers 416 views ### How to find eigenvalues of following block matrices? Is there a procedure to find the eigenvalues of A? ‎ $$A=\begin{bmatrix}X & I &&&&&&&&& 0\\I & 0 & P &&&&&&&&\\& P^t ... 0 votes 0 answers 192 views ### Upper bound on matrix perturbation such that all eigenvalues lie within the unit circle Consider the matrix$$N=\left[\matrix{\mathbb{I}_n-\epsilon L & X\\ \epsilon Y & Z}\right]$$where \epsilon>0 is a small positive parameter and Z is a square m\times m matrix with ... 7 votes 0 answers 517 views ### A rank inequality Suppose$$M := \begin{bmatrix} M_{11} & \cdots &M_{1d} \\ \vdots & \ddots & \vdots \\ M_{d1} & \cdots & M_{dd} \end{bmatrix}$$is a d \times d block matrix such that$$M_{... 3 votes 1 answer 2k views ### Bounds for eigenvalues of block matrix 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 ... 1 vote 0 answers 728 views ### Expressing a block matrix using Kronecker product [closed] Let$A=[a_{ij}]$be an$m \times m$matrix and$B$be a$m n \times m n$block diagonal matrix with$n \times n$diagonal blocks$B_1, B_2, \ldots, B_m$. I want to express the following block matrix$...
If I have a block matrix of the form $$M = \begin{pmatrix} A &B \\[6pt] -B & C \end{pmatrix}$$ and if $A$ is invertible I can write determinant in terms of the Schur ...