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Matrix theory is the study of matrices as concrete objects, rather than as abstract linear operators between vector spaces (whose study belongs to linear algebra). For instance, this involves matrix factorizations and decompositions, nonnegative matrices and Perron-Frobenius theory, Schur complements, structured and special matrices, matrix functions and equations.
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On ranks of matrices with tensor structure
Fix two $2^t$ length vector of form $p=\begin{bmatrix}u_1&v_1\end{bmatrix}\otimes\dots\otimes\begin{bmatrix}u_t&v_t\end{bmatrix}$ and $r=\begin{bmatrix}w_1&z_1\end{bmatrix}\otimes\dots\otimes\begin{bm …
2
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0
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Number of distinct rows and columns in a matrix with bounded number of entries
How many distinct rows and columns a real square matrix can have (at least in symmetric case) such that rank of matrix is $r$ and entries:
are from $\{-b,-b+1,\dots,0,\dots,b-1,b\}$?
are from $\{-b, …
6
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99
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What is this matrix decomposition called and does it exist always? - II
Given a rank $2r$ matrix $M\in\Bbb Q_{\geq0}^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank at most $r$ such that $M=M_+-M …
0
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1
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138
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On sum of matrices
Suppose we have a matrix $M\in\Bbb Q^{n\times n}$ with no $0$ elements and we write as sum of two matrices $M_1$ and $M_2$ on following constraint.
$M_{1,ij}=M_{ij}$ and $M_{2,ij}=0$ or $M_{1,ij}=0$ …
4
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291
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What is this matrix decomposition called and does it exist always?
Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?
Thou …
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A simple matrix multiplication query [closed]
The entries of $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a'&b'\\c'&d'\end{bmatrix}=\begin{bmatrix}aa'+bc'&ab'+bd'\\ca'+dc'&cb'+dd'\end{bmatrix}$ are curiously given by the entries of the com …
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When does the determinant distribute over addition?
When does $\det(A+B)=\det(A)+\det(B)$ hold?
I actually wonder if there is an easy answer for when $Per(A+B)=Per(A)+Per(B)$.