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LSpice
LSpice
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Let A$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Ker{Ker}$Let $A$ be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$$X^T$, $Y^T$ be their transposetransposes. Assume that $$p^dA^n\subset Im X+Im Y.$$$$p^dA^n\subset \Im X+\Im Y.$$

Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$$$p^{f(d)}A^n \subset \Im X^T+\Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Remark:

To answer the question of Luc Guyot in the commentscomments, note the following fact. This implies that, under the above assumption, at least the matrix $[X^T, Y^T]$ has rank $n$.

Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $Im X+Im Y=K^n$$\Im X+\Im Y=K^n$ if and only if $Ker X\cap Ker Y=0$$\Ker X\cap \Ker Y=0$.

Proof of the fact: We may assume that K$K$ is algebraically closed. Then, splitting the vector space as the direct sum of the characteristic spaces of $X$ (which are stable by $Y$), we may assume that $X$ has a single eigenvalue. If this eigenvalue is nonzero, then $Ker X=0$$\Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the same way with $Y$, we may assume that both $X$ and $Y$ are nilpotent. But then, since they commute, the assumption that $Im X+Im Y=K^n$$\Im X+\Im Y=K^n$ implies that $n=0$.

Let A be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$ be their transpose. Assume that $$p^dA^n\subset Im X+Im Y.$$

Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Remark:

To answer the question of Luc Guyot in the comments, note the following fact. This implies that, under the above assumption, at least the matrix $[X^T, Y^T]$ has rank $n$.

Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $Im X+Im Y=K^n$ if and only if $Ker X\cap Ker Y=0$.

Proof of the fact: We may assume that K is algebraically closed. Then, splitting the vector space as the direct sum of the characteristic spaces of $X$ (which are stable by $Y$), we may assume that $X$ has a single eigenvalue. If this eigenvalue is nonzero, then $Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the same way with $Y$, we may assume that both $X$ and $Y$ are nilpotent. But then, since they commute, the assumption that $Im X+Im Y=K^n$ implies that $n=0$.

$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Ker{Ker}$Let $A$ be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T$, $Y^T$ be their transposes. Assume that $$p^dA^n\subset \Im X+\Im Y.$$

Does this imply that $$p^{f(d)}A^n \subset \Im X^T+\Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Remark:

To answer the question of Luc Guyot in the comments, note the following fact. This implies that, under the above assumption, at least the matrix $[X^T, Y^T]$ has rank $n$.

Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $\Im X+\Im Y=K^n$ if and only if $\Ker X\cap \Ker Y=0$.

Proof of the fact: We may assume that $K$ is algebraically closed. Then, splitting the vector space as the direct sum of the characteristic spaces of $X$ (which are stable by $Y$), we may assume that $X$ has a single eigenvalue. If this eigenvalue is nonzero, then $\Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the same way with $Y$, we may assume that both $X$ and $Y$ are nilpotent. But then, since they commute, the assumption that $\Im X+\Im Y=K^n$ implies that $n=0$.

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Sylvain Brochard
Sylvain Brochard
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Let A be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$ be their transpose. Assume that $$p^dA^n\subset Im X+Im Y.$$ Does this imply

Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Remark:

To answer the question of Luc Guyot in the comments, note the following fact. This implies that, under the above assumption, at least the matrix $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where$[X^T, Y^T]$ has rank $f : \mathbb{N}\to \mathbb{N}$ is a function$n$.

Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $Im X+Im Y=K^n$ if and only if $Ker X\cap Ker Y=0$.

Proof of the fact: We may assume that does not depend onK is algebraically closed. Then, splitting the matricesvector space as the direct sum of the characteristic spaces of $X$ and (which are stable by $Y$), and not even on their sizewe may assume that $n$?$X$ has a single eigenvalue. If this eigenvalue is nonzero, then (It might depend on$Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the dvrsame way with $A$$Y$, if neededwe may assume that both $X$ and $Y$ are nilpotent.) But then, since they commute, the assumption that $Im X+Im Y=K^n$ implies that $n=0$.

Let A be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$ be their transpose. Assume that $$p^dA^n\subset Im X+Im Y.$$ Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Let A be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$ be their transpose. Assume that $$p^dA^n\subset Im X+Im Y.$$

Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)

Remark:

To answer the question of Luc Guyot in the comments, note the following fact. This implies that, under the above assumption, at least the matrix $[X^T, Y^T]$ has rank $n$.

Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $Im X+Im Y=K^n$ if and only if $Ker X\cap Ker Y=0$.

Proof of the fact: We may assume that K is algebraically closed. Then, splitting the vector space as the direct sum of the characteristic spaces of $X$ (which are stable by $Y$), we may assume that $X$ has a single eigenvalue. If this eigenvalue is nonzero, then $Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the same way with $Y$, we may assume that both $X$ and $Y$ are nilpotent. But then, since they commute, the assumption that $Im X+Im Y=K^n$ implies that $n=0$.

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Sylvain Brochard
Sylvain Brochard
  • 21
  • 3

Invariant factors and commuting matrices over a discrete valuation ring

Let A be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T, Y^T$ be their transpose. Assume that $$p^dA^n\subset Im X+Im Y.$$ Does this imply that $$p^{f(d)}A^n \subset Im X^T+Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)