Similar/Equivalent matrices and field extensions
2
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Let $F$ be a field and $F' supseteq F$ be a field extension. Suppose $A, B in M_n(F)$ are equivalent in $M_n(F')$, so there exist $P, Q in GL_n(F')$ such that $A = PBQ$. Is it necessarily true that $A$ and $B$ are equivalent in $M_n(F)$?
How about if "equivalent" is replaced with "similar"?
linear-algebra abstract-algebra
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
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add a comment |
2
$begingroup$
Let $F$ be a field and $F' supseteq F$ be a field extension. Suppose $A, B in M_n(F)$ are equivalent in $M_n(F')$, so there exist $P, Q in GL_n(F')$ such that $A = PBQ$. Is it necessarily true that $A$ and $B$ are equivalent in $M_n(F)$?
How about if "equivalent" is replaced with "similar"?
linear-algebra abstract-algebra
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
$endgroup$
1
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
1
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For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52
add a comment |
2
2
2
1
$begingroup$
Let $F$ be a field and $F' supseteq F$ be a field extension. Suppose $A, B in M_n(F)$ are equivalent in $M_n(F')$, so there exist $P, Q in GL_n(F')$ such that $A = PBQ$. Is it necessarily true that $A$ and $B$ are equivalent in $M_n(F)$?
How about if "equivalent" is replaced with "similar"?
linear-algebra abstract-algebra
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
$endgroup$
Let $F$ be a field and $F' supseteq F$ be a field extension. Suppose $A, B in M_n(F)$ are equivalent in $M_n(F')$, so there exist $P, Q in GL_n(F')$ such that $A = PBQ$. Is it necessarily true that $A$ and $B$ are equivalent in $M_n(F)$?
How about if "equivalent" is replaced with "similar"?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
asked Dec 4 '18 at 0:12
Jeremiah GoertzJeremiah Goertz
313
313
1
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
1
$begingroup$
For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52
add a comment |
1
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
1
$begingroup$
For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52
1
1
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
1
1
$begingroup$
For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52
add a comment |
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1
$begingroup$
For "similar", it is true and is a particular case of mathoverflow.net/questions/9162/… .
$endgroup$
– darij grinberg
Dec 4 '18 at 0:43
1
$begingroup$
For "equivalent" (I think the standard word for this is "congruent"), it is true because each matrix is equivalent to its rank normal form (and of course, the rank of a matrix does not change when we extend the field).
$endgroup$
– darij grinberg
Dec 4 '18 at 0:44
$begingroup$
Thanks for the replies! Very helpful.
$endgroup$
– Jeremiah Goertz
Dec 5 '18 at 1:52