Equivalence of matrices over $mathbb{Q}[x]$
$begingroup$
Someone asked a variant of this question recently (Equivalencity of $xI-A$). I ran across a suspiciously similar question today!
Let $A$ and $B$ be $n times n$ matrices over $mathbb{Q}$.
If $det(xI-A) = det(xI-B)$, can you say $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$? This is false and is answered here (Does equality of characteristic polynomials guarantee equivalence of matrices?)
If $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$, are $A$ and $B$ similar?
My suspicion is that (2.) is also false. If it were true, then we would have matrices $P,Q$ over $mathbb{Q}[x]$ such that $xI-A = P(xI-B)Q$. And I am not sure how to reduce matrices over $mathbb{Q}[x]$ to matrices over $mathbb{Q}$ to get the required similarity relation. Although, this seems to be a slightly stronger statement than $A$ and $B$ have the same characteristic polynomial.
Any help appreciated.
linear-algebra abstract-algebra
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
$endgroup$
add a comment |
$begingroup$
Someone asked a variant of this question recently (Equivalencity of $xI-A$). I ran across a suspiciously similar question today!
Let $A$ and $B$ be $n times n$ matrices over $mathbb{Q}$.
If $det(xI-A) = det(xI-B)$, can you say $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$? This is false and is answered here (Does equality of characteristic polynomials guarantee equivalence of matrices?)
If $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$, are $A$ and $B$ similar?
My suspicion is that (2.) is also false. If it were true, then we would have matrices $P,Q$ over $mathbb{Q}[x]$ such that $xI-A = P(xI-B)Q$. And I am not sure how to reduce matrices over $mathbb{Q}[x]$ to matrices over $mathbb{Q}$ to get the required similarity relation. Although, this seems to be a slightly stronger statement than $A$ and $B$ have the same characteristic polynomial.
Any help appreciated.
linear-algebra abstract-algebra
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
$endgroup$
$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50
add a comment |
$begingroup$
Someone asked a variant of this question recently (Equivalencity of $xI-A$). I ran across a suspiciously similar question today!
Let $A$ and $B$ be $n times n$ matrices over $mathbb{Q}$.
If $det(xI-A) = det(xI-B)$, can you say $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$? This is false and is answered here (Does equality of characteristic polynomials guarantee equivalence of matrices?)
If $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$, are $A$ and $B$ similar?
My suspicion is that (2.) is also false. If it were true, then we would have matrices $P,Q$ over $mathbb{Q}[x]$ such that $xI-A = P(xI-B)Q$. And I am not sure how to reduce matrices over $mathbb{Q}[x]$ to matrices over $mathbb{Q}$ to get the required similarity relation. Although, this seems to be a slightly stronger statement than $A$ and $B$ have the same characteristic polynomial.
Any help appreciated.
linear-algebra abstract-algebra
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
$endgroup$
Someone asked a variant of this question recently (Equivalencity of $xI-A$). I ran across a suspiciously similar question today!
Let $A$ and $B$ be $n times n$ matrices over $mathbb{Q}$.
If $det(xI-A) = det(xI-B)$, can you say $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$? This is false and is answered here (Does equality of characteristic polynomials guarantee equivalence of matrices?)
If $xI-A$ and $xI-B$ are equivalent over $mathbb{Q}[x]$, are $A$ and $B$ similar?
My suspicion is that (2.) is also false. If it were true, then we would have matrices $P,Q$ over $mathbb{Q}[x]$ such that $xI-A = P(xI-B)Q$. And I am not sure how to reduce matrices over $mathbb{Q}[x]$ to matrices over $mathbb{Q}$ to get the required similarity relation. Although, this seems to be a slightly stronger statement than $A$ and $B$ have the same characteristic polynomial.
Any help appreciated.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
edited Dec 16 '18 at 3:33
Pentaki
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
asked Dec 16 '18 at 2:31
PentakiPentaki
1105
1105
$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50
add a comment |
$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50
$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50
$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50
add a comment |
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$begingroup$
2) is true. $A-lambda$ and $B-lambda$ being $lambda$-equivalent is the necessary and sufficient condition for $A,B$ being similar.
$endgroup$
– Vim
Dec 16 '18 at 3:50