Counting Characteristic Polynomials?

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Question: Let $M_n$ be the set of all square $n times n$ $(0,1)$-matrices. I write $$mathbfchar(M_n)=chi(X)text $$ How do I compute the order of $mathbfchar(M_n)$ ?




For example: $M_2$ has $16$ distinct matrices with $6$ distinct characteristic polynomials:
$$mathbfchar(M_2)=X^2,X^2-X,X^2-2X+1,X^2-X-1,X^2-2X,X^2-1$$



I know that similar matrices have the same characteristic polynomial but the converse is false. I am not certain the question is equivalent to counting similar matrices ?




linear-algebra abstract-algebra combinatorics matrices polynomials




share|cite|improve this question

















  • 1




    Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
    – Christopher
    Jul 19 at 16:12






  • 1




    I will edit and correct the question.
    – Anthony Hernandez
    Jul 19 at 16:17






  • 1




    This is OEISA272661. No formula is given.
    – Jair Taylor
    Jul 19 at 18:57






  • 1




    The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
    – Jair Taylor
    Jul 19 at 19:04














up vote
1
down vote

favorite
1













Question: Let $M_n$ be the set of all square $n times n$ $(0,1)$-matrices. I write $$mathbfchar(M_n)=chi(X)text $$ How do I compute the order of $mathbfchar(M_n)$ ?




For example: $M_2$ has $16$ distinct matrices with $6$ distinct characteristic polynomials:
$$mathbfchar(M_2)=X^2,X^2-X,X^2-2X+1,X^2-X-1,X^2-2X,X^2-1$$



I know that similar matrices have the same characteristic polynomial but the converse is false. I am not certain the question is equivalent to counting similar matrices ?




linear-algebra abstract-algebra combinatorics matrices polynomials




share|cite|improve this question

















  • 1




    Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
    – Christopher
    Jul 19 at 16:12






  • 1




    I will edit and correct the question.
    – Anthony Hernandez
    Jul 19 at 16:17






  • 1




    This is OEISA272661. No formula is given.
    – Jair Taylor
    Jul 19 at 18:57






  • 1




    The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
    – Jair Taylor
    Jul 19 at 19:04












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






Question: Let $M_n$ be the set of all square $n times n$ $(0,1)$-matrices. I write $$mathbfchar(M_n)=chi(X)text $$ How do I compute the order of $mathbfchar(M_n)$ ?




For example: $M_2$ has $16$ distinct matrices with $6$ distinct characteristic polynomials:
$$mathbfchar(M_2)=X^2,X^2-X,X^2-2X+1,X^2-X-1,X^2-2X,X^2-1$$



I know that similar matrices have the same characteristic polynomial but the converse is false. I am not certain the question is equivalent to counting similar matrices ?




linear-algebra abstract-algebra combinatorics matrices polynomials




share|cite|improve this question














Question: Let $M_n$ be the set of all square $n times n$ $(0,1)$-matrices. I write $$mathbfchar(M_n)=chi(X)text $$ How do I compute the order of $mathbfchar(M_n)$ ?




For example: $M_2$ has $16$ distinct matrices with $6$ distinct characteristic polynomials:
$$mathbfchar(M_2)=X^2,X^2-X,X^2-2X+1,X^2-X-1,X^2-2X,X^2-1$$



I know that similar matrices have the same characteristic polynomial but the converse is false. I am not certain the question is equivalent to counting similar matrices ?





linear-algebra abstract-algebra combinatorics matrices polynomials





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 17:38
























asked Jul 19 at 15:47









Anthony Hernandez

1,224619




1,224619







  • 1




    Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
    – Christopher
    Jul 19 at 16:12






  • 1




    I will edit and correct the question.
    – Anthony Hernandez
    Jul 19 at 16:17






  • 1




    This is OEISA272661. No formula is given.
    – Jair Taylor
    Jul 19 at 18:57






  • 1




    The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
    – Jair Taylor
    Jul 19 at 19:04












  • 1




    Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
    – Christopher
    Jul 19 at 16:12






  • 1




    I will edit and correct the question.
    – Anthony Hernandez
    Jul 19 at 16:17






  • 1




    This is OEISA272661. No formula is given.
    – Jair Taylor
    Jul 19 at 18:57






  • 1




    The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
    – Jair Taylor
    Jul 19 at 19:04







1




1




Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
– Christopher
Jul 19 at 16:12




Is the $n$ in $M_n$ meant to be the same as the $n$ in $q = p^n$? Along similar lines, it's not clear what field you are working over in your example: $|M_2| = 16$ would imply you're working over $Bbb F_2$ but your list of characteristic polynomials includes $X^2 - 2X$ which equals $X^2$ in a field of characteristic $2$.
– Christopher
Jul 19 at 16:12




1




1




I will edit and correct the question.
– Anthony Hernandez
Jul 19 at 16:17




I will edit and correct the question.
– Anthony Hernandez
Jul 19 at 16:17




1




1




This is OEISA272661. No formula is given.
– Jair Taylor
Jul 19 at 18:57




This is OEISA272661. No formula is given.
– Jair Taylor
Jul 19 at 18:57




1




1




The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
– Jair Taylor
Jul 19 at 19:04




The slides linked to in the OEIS entry seem helpful also (with some nice visuals.)
– Jair Taylor
Jul 19 at 19:04










1 Answer
1






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up vote
1
down vote













Edit: as pointed out in the comments, I had misunderstood the question



For any monic polynomial $p$ of degree $n$ over a field $K$, one can construct its companion matrix which is an $n times n$ matrix over $K$ with characteristic polynomial $p$. Therefore counting the number of possible characteristic polynomials of $n times n$ matrices over $K$ is the same as counting the number of possible monic polynomials of degree $n$ over $K$, i.e. $|K|^n$.






share|cite|improve this answer























  • Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
    – Anthony Hernandez
    Jul 19 at 17:06






  • 2




    The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
    – Jair Taylor
    Jul 19 at 17:44











  • Yes @JairTaylor
    – Anthony Hernandez
    Jul 19 at 17:45










  • Ah, I see. Thank you @JairTaylor
    – Christopher
    Jul 19 at 17:46










  • @JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
    – Anthony Hernandez
    Jul 19 at 19:31










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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
1
down vote













Edit: as pointed out in the comments, I had misunderstood the question



For any monic polynomial $p$ of degree $n$ over a field $K$, one can construct its companion matrix which is an $n times n$ matrix over $K$ with characteristic polynomial $p$. Therefore counting the number of possible characteristic polynomials of $n times n$ matrices over $K$ is the same as counting the number of possible monic polynomials of degree $n$ over $K$, i.e. $|K|^n$.






share|cite|improve this answer























  • Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
    – Anthony Hernandez
    Jul 19 at 17:06






  • 2




    The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
    – Jair Taylor
    Jul 19 at 17:44











  • Yes @JairTaylor
    – Anthony Hernandez
    Jul 19 at 17:45










  • Ah, I see. Thank you @JairTaylor
    – Christopher
    Jul 19 at 17:46










  • @JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
    – Anthony Hernandez
    Jul 19 at 19:31














up vote
1
down vote













Edit: as pointed out in the comments, I had misunderstood the question



For any monic polynomial $p$ of degree $n$ over a field $K$, one can construct its companion matrix which is an $n times n$ matrix over $K$ with characteristic polynomial $p$. Therefore counting the number of possible characteristic polynomials of $n times n$ matrices over $K$ is the same as counting the number of possible monic polynomials of degree $n$ over $K$, i.e. $|K|^n$.






share|cite|improve this answer























  • Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
    – Anthony Hernandez
    Jul 19 at 17:06






  • 2




    The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
    – Jair Taylor
    Jul 19 at 17:44











  • Yes @JairTaylor
    – Anthony Hernandez
    Jul 19 at 17:45










  • Ah, I see. Thank you @JairTaylor
    – Christopher
    Jul 19 at 17:46










  • @JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
    – Anthony Hernandez
    Jul 19 at 19:31












up vote
1
down vote










up vote
1
down vote









Edit: as pointed out in the comments, I had misunderstood the question



For any monic polynomial $p$ of degree $n$ over a field $K$, one can construct its companion matrix which is an $n times n$ matrix over $K$ with characteristic polynomial $p$. Therefore counting the number of possible characteristic polynomials of $n times n$ matrices over $K$ is the same as counting the number of possible monic polynomials of degree $n$ over $K$, i.e. $|K|^n$.






share|cite|improve this answer















Edit: as pointed out in the comments, I had misunderstood the question



For any monic polynomial $p$ of degree $n$ over a field $K$, one can construct its companion matrix which is an $n times n$ matrix over $K$ with characteristic polynomial $p$. Therefore counting the number of possible characteristic polynomials of $n times n$ matrices over $K$ is the same as counting the number of possible monic polynomials of degree $n$ over $K$, i.e. $|K|^n$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 19 at 17:47


























answered Jul 19 at 16:45









Christopher

4,8251224




4,8251224











  • Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
    – Anthony Hernandez
    Jul 19 at 17:06






  • 2




    The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
    – Jair Taylor
    Jul 19 at 17:44











  • Yes @JairTaylor
    – Anthony Hernandez
    Jul 19 at 17:45










  • Ah, I see. Thank you @JairTaylor
    – Christopher
    Jul 19 at 17:46










  • @JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
    – Anthony Hernandez
    Jul 19 at 19:31
















  • Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
    – Anthony Hernandez
    Jul 19 at 17:06






  • 2




    The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
    – Jair Taylor
    Jul 19 at 17:44











  • Yes @JairTaylor
    – Anthony Hernandez
    Jul 19 at 17:45










  • Ah, I see. Thank you @JairTaylor
    – Christopher
    Jul 19 at 17:46










  • @JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
    – Anthony Hernandez
    Jul 19 at 19:31















Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
– Anthony Hernandez
Jul 19 at 17:06




Hmm so then my example is wrong ? We should only have $4$ distinct characteristic polynomials ? Also here
– Anthony Hernandez
Jul 19 at 17:06




2




2




The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
– Jair Taylor
Jul 19 at 17:44





The OP not considering matrices over a finite field. He's considering matrices (over the rationals, say) that happen to have only 0's and 1's.
– Jair Taylor
Jul 19 at 17:44













Yes @JairTaylor
– Anthony Hernandez
Jul 19 at 17:45




Yes @JairTaylor
– Anthony Hernandez
Jul 19 at 17:45












Ah, I see. Thank you @JairTaylor
– Christopher
Jul 19 at 17:46




Ah, I see. Thank you @JairTaylor
– Christopher
Jul 19 at 17:46












@JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
– Anthony Hernandez
Jul 19 at 19:31




@JairTaylor can you answerize your comment so I can throw you some points. And it appears the formula $mathbfchar(M_n)$ is not know.
– Anthony Hernandez
Jul 19 at 19:31












 

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