Invariant Factors of $Ain M_{4times 4}(mathbb{Q})$. Announcing the arrival of Valued Associate...
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Invariant Factors of $Ain M_{4times 4}(mathbb{Q})$.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Invariant Factors of a Module over R[x]Finding invariant factors of finitely generated Abelian groupConfusion regarding finding invariant factors of a matrix.Classifying similarity classes of matrices based on number of invariant subspaces of a given dimensionConfusion with Smith normal form and rational canonical form.What is the purpose of these extra steps in the algorithm for converting to rational canonical form?Invariant factors of a module over a polynomial-ringFinding invariant factors of power matrixSmiths normal form is similar to $xI-A.$Invariant Factors Example
$begingroup$
Suppose that $A$ is a $4times 4$ matrix over $mathbb{Q}$ which satisfies $(A^2+I)^2=0$. What are the possibilities for the invariant factors of $A$? Describe $A$ up to similarity.
Im totally lost on this one. I have a feeling I may need to use something like the Rational Canonical Form. Any hints on this are much appreciated.
abstract-algebra modules
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
$endgroup$
add a comment |
$begingroup$
Suppose that $A$ is a $4times 4$ matrix over $mathbb{Q}$ which satisfies $(A^2+I)^2=0$. What are the possibilities for the invariant factors of $A$? Describe $A$ up to similarity.
Im totally lost on this one. I have a feeling I may need to use something like the Rational Canonical Form. Any hints on this are much appreciated.
abstract-algebra modules
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
$endgroup$
$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54
add a comment |
$begingroup$
Suppose that $A$ is a $4times 4$ matrix over $mathbb{Q}$ which satisfies $(A^2+I)^2=0$. What are the possibilities for the invariant factors of $A$? Describe $A$ up to similarity.
Im totally lost on this one. I have a feeling I may need to use something like the Rational Canonical Form. Any hints on this are much appreciated.
abstract-algebra modules
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
$endgroup$
Suppose that $A$ is a $4times 4$ matrix over $mathbb{Q}$ which satisfies $(A^2+I)^2=0$. What are the possibilities for the invariant factors of $A$? Describe $A$ up to similarity.
Im totally lost on this one. I have a feeling I may need to use something like the Rational Canonical Form. Any hints on this are much appreciated.
abstract-algebra modules
abstract-algebra modules
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
asked Mar 26 at 11:15
Joe Man AnalysisJoe Man Analysis
69319
69319
$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54
add a comment |
$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54
$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54
$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54
add a comment |
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$begingroup$
The minimal polynomial of $A$ is its invariant factor of largest degree. It also must divide $(x^2+1)^2$ and have rational coefficients. So, it could be $(x^2+1)^2$ or $x^2+1$. In the first case, the companion matrix is already $4times4$. In the second case we need more invariant factors. They must divide $x^2+1$, be monic, and non-constant. That would be another $x^2+1$. The side length of the two companion matrices of these already add up to $4$.
$endgroup$
– user647486
Mar 26 at 11:54