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Math.TechQA.Club
2026-02-23 11:25:30
630
Views
Smith Normal Form
Published on
23 Feb 2026 - 11:25
#linear-algebra
#abstract-algebra
#matrices
#diagonalization
#smith-normal-form
620
Views
Smith Normal Form of this polynomial non square matrix in $\mathbb Q[t]$
Published on
23 Feb 2026 - 11:33
#abstract-algebra
#matrices
#polynomials
#divisibility
#smith-normal-form
1.4k
Views
Finding Jordan normal form using Smith normal form
Published on
23 Feb 2026 - 11:25
#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
23 Feb 2026 - 11:26
#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
23 Feb 2026 - 11:28
#linear-algebra
#principal-ideal-domains
#smith-normal-form
1.4k
Views
Smith Normal Form
Published on
23 Feb 2026 - 11:25
#linear-algebra
#matrices
#ring-theory
#smith-normal-form
27.8k
Views
Computing the Smith Normal Form
Published on
23 Feb 2026 - 11:25
#matrices
#finite-groups
#abelian-groups
#matrix-decomposition
#smith-normal-form
1.8k
Views
Determining the Smith Normal Form
Published on
23 Feb 2026 - 11:25
#matrices
#abelian-groups
#smith-normal-form
225
Views
Smith normal form of powers of a matrix
Published on
23 Feb 2026 - 11:30
#linear-algebra
#abstract-algebra
#matrices
#smith-normal-form
6.6k
Views
Smith normal form of a polynomial matrix
Published on
23 Feb 2026 - 11:27
#matrices
#polynomials
#control-theory
#smith-normal-form
89
Views
Finding Smith Normal Form of the Inverse of a Matrix
Published on
23 Feb 2026 - 11:30
#linear-algebra
#matrices
#smith-normal-form
128
Views
Is $\mathbb Z$ isomorphic to a polynomial ring over a field?
Published on
23 Feb 2026 - 11:28
#abstract-algebra
#smith-normal-form
481
Views
Smith normal form and elementary divisors
Published on
23 Feb 2026 - 11:27
#linear-algebra
#matrices
#modules
#free-modules
#smith-normal-form
110
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
23 Feb 2026 - 11:33
#linear-algebra
#abstract-algebra
#matrices
#free-modules
#smith-normal-form
21
Views
Taking quotients over isometric groups and determining the cokernel of transformations
Published on
23 Feb 2026 - 11:29
#quotient-group
#free-modules
#smith-normal-form
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