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15
Math.TechQA.Club
2017-06-12 19:17:14
629
Views
Smith Normal Form
Published on
12 Jun 2017 - 19:17
#linear-algebra
#abstract-algebra
#matrices
#diagonalization
#smith-normal-form
619
Views
Smith Normal Form of this polynomial non square matrix in $\mathbb Q[t]$
Published on
23 Jun 2017 - 10:36
#abstract-algebra
#matrices
#polynomials
#divisibility
#smith-normal-form
1.4k
Views
Finding Jordan normal form using Smith normal form
Published on
24 Nov 2017 - 18:48
#linear-algebra
#modules
#jordan-normal-form
#vector-space-isomorphism
#smith-normal-form
2.4k
Views
Finding the correct pivot in Smith normal form
Published on
27 Jul 2011 - 20:28
#linear-algebra
#abstract-algebra
#smith-normal-form
3.4k
Views
How do I get this matrix in Smith Normal Form? And, is Smith Normal Form unique?
Published on
30 Oct 2011 - 0:57
#linear-algebra
#principal-ideal-domains
#smith-normal-form
1.4k
Views
Smith Normal Form
Published on
15 Mar 2012 - 23:07
#linear-algebra
#matrices
#ring-theory
#smith-normal-form
27.8k
Views
Computing the Smith Normal Form
Published on
17 Apr 2012 - 19:05
#matrices
#finite-groups
#abelian-groups
#matrix-decomposition
#smith-normal-form
1.8k
Views
Determining the Smith Normal Form
Published on
13 May 2012 - 13:47
#matrices
#abelian-groups
#smith-normal-form
224
Views
Smith normal form of powers of a matrix
Published on
19 Aug 2012 - 1:29
#linear-algebra
#abstract-algebra
#matrices
#smith-normal-form
6.6k
Views
Smith normal form of a polynomial matrix
Published on
28 Oct 2012 - 16:38
#matrices
#polynomials
#control-theory
#smith-normal-form
88
Views
Finding Smith Normal Form of the Inverse of a Matrix
Published on
18 Nov 2020 - 19:40
#linear-algebra
#matrices
#smith-normal-form
127
Views
Is $\mathbb Z$ isomorphic to a polynomial ring over a field?
Published on
15 Dec 2020 - 17:37
#abstract-algebra
#smith-normal-form
480
Views
Smith normal form and elementary divisors
Published on
20 Dec 2020 - 17:45
#linear-algebra
#matrices
#modules
#free-modules
#smith-normal-form
109
Views
If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is a module homomorphism, then $\mathbb{Z}^m/\operatorname{im}(f)$ is a finite abelian group
Published on
20 Dec 2020 - 18:05
#linear-algebra
#abstract-algebra
#matrices
#free-modules
#smith-normal-form
20
Views
Taking quotients over isometric groups and determining the cokernel of transformations
Published on
25 Jan 2021 - 10:54
#quotient-group
#free-modules
#smith-normal-form
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