Equivalencity of $xI-A$
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Let $A$ and $B$ be two matrices over $mathbb Q[x]$. What is the relation be the the conditions that (1) $det(xI-A) = det (xI-B)$ and that (2) $A$ and $B$ are equivalent, that is, there exists invertible matrices $P$ and $Q$ over $mathbb Q[x]$ such that $A = PBQ$.
Do these two conditions imply each other? If so, what is a proof? If one does not imply the other, why not?
linear-algebra abstract-algebra
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
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|
show 1 more comment
$begingroup$
Let $A$ and $B$ be two matrices over $mathbb Q[x]$. What is the relation be the the conditions that (1) $det(xI-A) = det (xI-B)$ and that (2) $A$ and $B$ are equivalent, that is, there exists invertible matrices $P$ and $Q$ over $mathbb Q[x]$ such that $A = PBQ$.
Do these two conditions imply each other? If so, what is a proof? If one does not imply the other, why not?
linear-algebra abstract-algebra
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
$endgroup$
$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
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– Arnaud D.
Dec 14 '18 at 16:36
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What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
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@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
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@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34
|
show 1 more comment
$begingroup$
Let $A$ and $B$ be two matrices over $mathbb Q[x]$. What is the relation be the the conditions that (1) $det(xI-A) = det (xI-B)$ and that (2) $A$ and $B$ are equivalent, that is, there exists invertible matrices $P$ and $Q$ over $mathbb Q[x]$ such that $A = PBQ$.
Do these two conditions imply each other? If so, what is a proof? If one does not imply the other, why not?
linear-algebra abstract-algebra
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
$endgroup$
Let $A$ and $B$ be two matrices over $mathbb Q[x]$. What is the relation be the the conditions that (1) $det(xI-A) = det (xI-B)$ and that (2) $A$ and $B$ are equivalent, that is, there exists invertible matrices $P$ and $Q$ over $mathbb Q[x]$ such that $A = PBQ$.
Do these two conditions imply each other? If so, what is a proof? If one does not imply the other, why not?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
edited Dec 14 '18 at 17:52
LinearGuy
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
asked Dec 14 '18 at 16:30
LinearGuyLinearGuy
1511
1511
$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:36
$begingroup$
What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
$begingroup$
@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
$begingroup$
@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34
|
show 1 more comment
$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:36
$begingroup$
What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
$begingroup$
@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
$begingroup$
@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34
$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:36
$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:36
$begingroup$
What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
$begingroup$
What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
$begingroup$
@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
$begingroup$
@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
$begingroup$
@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
No relations.
Example 1. Let $A=begin{pmatrix} 0 &0 \ 0 & 0end{pmatrix}$ and
$B=begin{pmatrix} 0 &1 \ 0 & 0end{pmatrix}$ They have the same characteristic polynomial ($det(Ix-A)=det(Ix-B)=x^2$) but the only matrix equivalent to $A$ is $A$ itself.
Example 2.
Let $A=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$B=begin{pmatrix} 1 &0 \ 0 & 2end{pmatrix}$
They have different characteristic polynomials but
if $P=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$Q=begin{pmatrix} 1 &0 \ 0 & 1/2end{pmatrix}$
then $PBQ=A$.
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
$endgroup$
add a comment |
$begingroup$
None of the condition implies the other.
Take B to be the $0$ matrix and A to be the $0$ matrix with a nonzero element in the upper right corner. det(A-xI)=det(B-xI) but there are no invertible matrices P and Q as you require.
For the second condition to imply the first any two invertible matrices would have to have the same characteristic poly which is absurd
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No relations.
Example 1. Let $A=begin{pmatrix} 0 &0 \ 0 & 0end{pmatrix}$ and
$B=begin{pmatrix} 0 &1 \ 0 & 0end{pmatrix}$ They have the same characteristic polynomial ($det(Ix-A)=det(Ix-B)=x^2$) but the only matrix equivalent to $A$ is $A$ itself.
Example 2.
Let $A=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$B=begin{pmatrix} 1 &0 \ 0 & 2end{pmatrix}$
They have different characteristic polynomials but
if $P=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$Q=begin{pmatrix} 1 &0 \ 0 & 1/2end{pmatrix}$
then $PBQ=A$.
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
$endgroup$
add a comment |
$begingroup$
No relations.
Example 1. Let $A=begin{pmatrix} 0 &0 \ 0 & 0end{pmatrix}$ and
$B=begin{pmatrix} 0 &1 \ 0 & 0end{pmatrix}$ They have the same characteristic polynomial ($det(Ix-A)=det(Ix-B)=x^2$) but the only matrix equivalent to $A$ is $A$ itself.
Example 2.
Let $A=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$B=begin{pmatrix} 1 &0 \ 0 & 2end{pmatrix}$
They have different characteristic polynomials but
if $P=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$Q=begin{pmatrix} 1 &0 \ 0 & 1/2end{pmatrix}$
then $PBQ=A$.
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
$endgroup$
add a comment |
$begingroup$
No relations.
Example 1. Let $A=begin{pmatrix} 0 &0 \ 0 & 0end{pmatrix}$ and
$B=begin{pmatrix} 0 &1 \ 0 & 0end{pmatrix}$ They have the same characteristic polynomial ($det(Ix-A)=det(Ix-B)=x^2$) but the only matrix equivalent to $A$ is $A$ itself.
Example 2.
Let $A=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$B=begin{pmatrix} 1 &0 \ 0 & 2end{pmatrix}$
They have different characteristic polynomials but
if $P=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$Q=begin{pmatrix} 1 &0 \ 0 & 1/2end{pmatrix}$
then $PBQ=A$.
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
$endgroup$
No relations.
Example 1. Let $A=begin{pmatrix} 0 &0 \ 0 & 0end{pmatrix}$ and
$B=begin{pmatrix} 0 &1 \ 0 & 0end{pmatrix}$ They have the same characteristic polynomial ($det(Ix-A)=det(Ix-B)=x^2$) but the only matrix equivalent to $A$ is $A$ itself.
Example 2.
Let $A=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$B=begin{pmatrix} 1 &0 \ 0 & 2end{pmatrix}$
They have different characteristic polynomials but
if $P=begin{pmatrix} 1 &0 \ 0 & 1end{pmatrix}$ and
$Q=begin{pmatrix} 1 &0 \ 0 & 1/2end{pmatrix}$
then $PBQ=A$.
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
answered Dec 14 '18 at 18:04
user126154user126154
5,376816
5,376816
add a comment |
add a comment |
$begingroup$
None of the condition implies the other.
Take B to be the $0$ matrix and A to be the $0$ matrix with a nonzero element in the upper right corner. det(A-xI)=det(B-xI) but there are no invertible matrices P and Q as you require.
For the second condition to imply the first any two invertible matrices would have to have the same characteristic poly which is absurd
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
$endgroup$
add a comment |
$begingroup$
None of the condition implies the other.
Take B to be the $0$ matrix and A to be the $0$ matrix with a nonzero element in the upper right corner. det(A-xI)=det(B-xI) but there are no invertible matrices P and Q as you require.
For the second condition to imply the first any two invertible matrices would have to have the same characteristic poly which is absurd
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
$endgroup$
add a comment |
$begingroup$
None of the condition implies the other.
Take B to be the $0$ matrix and A to be the $0$ matrix with a nonzero element in the upper right corner. det(A-xI)=det(B-xI) but there are no invertible matrices P and Q as you require.
For the second condition to imply the first any two invertible matrices would have to have the same characteristic poly which is absurd
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
$endgroup$
None of the condition implies the other.
Take B to be the $0$ matrix and A to be the $0$ matrix with a nonzero element in the upper right corner. det(A-xI)=det(B-xI) but there are no invertible matrices P and Q as you require.
For the second condition to imply the first any two invertible matrices would have to have the same characteristic poly which is absurd
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
answered Dec 14 '18 at 18:08
Sorin TircSorin Tirc
1,810213
1,810213
add a comment |
add a comment |
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$begingroup$
This looks interesting, but it would be nice if you could add some context to your question. For example, what you think of the problem, how you've encountered it, what you've tried, where you got stuck...
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:36
$begingroup$
What do you know about equivalent matrices? For example, can you see that two matrices are equivalent if and only if they have the same rank?
$endgroup$
– Zach Langley
Dec 14 '18 at 16:42
$begingroup$
@ZachLangley Are you sure that this would still be true for matrices over $mathbb{Q}[x]$ (which is not a field)?
$endgroup$
– Arnaud D.
Dec 14 '18 at 16:45
$begingroup$
@ArnaudD. Ah, no. I am not.
$endgroup$
– Zach Langley
Dec 14 '18 at 16:53
$begingroup$
Similarity implies same rank but the other implication is true in general?
$endgroup$
– orange
Dec 14 '18 at 17:34