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Results tagged with matrices
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user 2530
Questions where the notion matrix has an important or crucial role (for the latter, note the tag matrix-theory for potential use). Matrices appear in various parts of mathematics, and this tag is typically combined with other tags to make the general subject clear, such as an appropriate top-level tag ra.rings-and-algebras, co.combinatorics, etc. and other tags that might be applicable. There are also several more specialized tags concerning matrices.
19
votes
Jacobi's equality between complementary minors of inverse matrices
All matrices that appear in the
following are matrices over $\mathbb{K}$.
Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. …
- 33.8k
14
votes
Cayley-Hamilton revisited
Am I missing something or is Ilya Bogdanov's elimination of $A_0$ trick more or less a proof in itself?
Assume that $f\left(B\right) = 0_n$. Then, $0_n = f\left(B\right) = A_kB^k + A_{k-1}B^{k-1} + . …
- 33.8k
14
votes
3
answers
1k
views
"Conjugacy rank" of two matrices over field extension
\end{align}
We can call this number $\rho_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$. … matrices $A$ and $B$ are conjugate to each other in the ring $\operatorname{M}_{n}\left( S\right)$.) …
- 33.8k
11
votes
Accepted
A question about symmetric matrix
The Birkhoff-von Neumann theorem yields that it is a convex combination of permutation matrices. Take any permutation matrix which enters into this combination with a nonzero coefficient. …
- 33.8k
9
votes
Vandermonde matrix is totally positive
The minors of a Vandermonde matrix are known as alternants. Their positivity follows from the fact that they are products of the Vandermonde determinant of the variables involved with a Schur function …
- 33.8k
8
votes
Accepted
Why does this antisymmetric product factor out a determinant?
This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime}
=N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left( n-1\right)
\times\left( n-1\right) $-matrices). …
- 33.8k
8
votes
Accepted
Is the componentwise square-root of a positive-definite matrix also pos.-def.?
Counterexample:
$B=\left(\begin{array}{ccc} 1 & 0 & 1 \\\\ 0 & 1 & 1 \\\\ 1 & 1 & \sqrt3 \end{array}\right)$.
This is for $n=3$, and easily extends to all $n\geq 3$.
For $n=2$ it holds, though.
- 33.8k
8
votes
1
answer
490
views
ad (A^n) is a polynomial in ad A ?
Let $k$ be a field and $n$ a nonnegative integer. For any matrix $U\in\mathrm{M}_n\left(k\right)$, let $\mathrm{ad} U$ denote the map $\mathrm{M}_n\left(k\right)\to \mathrm{M}_n\left(k\right),\ V\maps …
- 33.8k
8
votes
0
answers
296
views
How to check two matrices for similitude over $\mathbb{Z}$?
I am particularly interested in two special matrices, more on which below. … Indeed, a proof of this fact has been
sketched in on similar matrices and polynomial matrices for any commutative ring
instead of $\mathbb{Z}$. …
- 33.8k
8
votes
3
answers
418
views
Smallest faithful representation of an upper-triangular matrix quotient
Let $J$ be the subset of $F^{n \leq n}$ that consists of all such matrices that satisfy $a_{i,j} = 0$ for all $\left(i,j\right) \neq \left(1,n\right)$. … Thus, a quotient $F$-algebra $F^{n \leq n} / J$ is defined (and its elements can be thought of as upper-triangular matrices whose northeasternmost entry is undetermined).
Question. …
- 33.8k
7
votes
Product of Positive Matrices
Let $A\in k^{n\times n}$ and $B\in k^{n\times n}$ be two symmetric nonnegative-definite matrices. Then, $\operatorname{Tr}\left(AB\right)\geq 0$.
Proof of Corollary 2. … Consider the Kronecker product $A\otimes B\in k^{n^2\times n^2}$ of the two matrices $A$ and $B$. …
- 33.8k
7
votes
1
answer
851
views
Invariant quadratic forms of irreducible representations
Let $G$ be a finite group, and $k$ be a field of characteristic zero (not necessarily algebraically closed!). Let $\rho : G \to \mathrm{End}_k \left(k^n\right)$ be a irreducible representation of $G$ …
- 33.8k
6
votes
Matrix decomposition the other way
The decomposition should work if and only if the minimal polynomial of the matrix can be factored into pairwise non-proportional linear factors.
Let me tell you how algebraists think about this (depe …
- 33.8k
5
votes
Accepted
On a determinantal equality
Let me prove a slightly more general claim.
Theorem 1. Let $\mathbb{K}$ be a commutative ring. Let $R$ be a set. Let
$n\in\mathbb{N}$ (where $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $). Set
…
- 33.8k
5
votes
Accepted
Determinants: periodic entries $0,1,2,3$
Yes, it is true. More generally, the entries $1,2,3,0$ can be replaced by arbitrary numbers $a,b,c,d$, in which case the determinant of $M_n$ can be computed in terms of the four numbers $u = d-b$, $v …
- 33.8k