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Counting the powers of a companion matrix that possess nonzero leading principal minors


Universal equivalence to P-matricesDeducing that a matrix is indefinite using only its leading principal minorsInvertible matrices, permutations and leading principal minorsGiven the roots of polynomial over finite field, what is the count of its distinct nonzero coefficients?Looking for a proof of a known theorem involving integral quadratic formsgeneralization of positive-definite matrices to matrices over finite fieldsShow that matrix $A_k$ is also symmetric and positive definite.Leading principal minors criterion for positive semi-definite matricesIs the set of real matrices diagonalizable in $M_n(mathbb C)$ dense in the set of block like companion matrices?Unramified primes inside $mathbbF_q(t)$













1












$begingroup$


Let $p$ be prime, let $A$ be the $ntimes n$ companion matrix associated to a primitive polynomial $f(x)$ of degree $n$ with coefficients in $mathbb Z_p$, and let $T=A^jmid 1leq jleq p^n-1subseteq mathbb Z_p^ntimes n$. Note that $T$ can be identified with the nonzero elements of $GF(p^n)$. Suppose $S$ is the subset of $T$ consisting of matrices possessing the property that every leading principal minor is nonzero. A couple of questions: (1) What is the cardinality of $S$? and (2) Does $S$ have some special meaning as a subset of $GF(p^n)$ or otherwise? Perhaps (1) depends on $f(x)$.



Thanks for your thoughts!










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    1












    $begingroup$


    Let $p$ be prime, let $A$ be the $ntimes n$ companion matrix associated to a primitive polynomial $f(x)$ of degree $n$ with coefficients in $mathbb Z_p$, and let $T=A^jmid 1leq jleq p^n-1subseteq mathbb Z_p^ntimes n$. Note that $T$ can be identified with the nonzero elements of $GF(p^n)$. Suppose $S$ is the subset of $T$ consisting of matrices possessing the property that every leading principal minor is nonzero. A couple of questions: (1) What is the cardinality of $S$? and (2) Does $S$ have some special meaning as a subset of $GF(p^n)$ or otherwise? Perhaps (1) depends on $f(x)$.



    Thanks for your thoughts!










    share|cite|improve this question







    New contributor




    Murphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      1












      1








      1





      $begingroup$


      Let $p$ be prime, let $A$ be the $ntimes n$ companion matrix associated to a primitive polynomial $f(x)$ of degree $n$ with coefficients in $mathbb Z_p$, and let $T=A^jmid 1leq jleq p^n-1subseteq mathbb Z_p^ntimes n$. Note that $T$ can be identified with the nonzero elements of $GF(p^n)$. Suppose $S$ is the subset of $T$ consisting of matrices possessing the property that every leading principal minor is nonzero. A couple of questions: (1) What is the cardinality of $S$? and (2) Does $S$ have some special meaning as a subset of $GF(p^n)$ or otherwise? Perhaps (1) depends on $f(x)$.



      Thanks for your thoughts!










      share|cite|improve this question







      New contributor




      Murphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $p$ be prime, let $A$ be the $ntimes n$ companion matrix associated to a primitive polynomial $f(x)$ of degree $n$ with coefficients in $mathbb Z_p$, and let $T=A^jmid 1leq jleq p^n-1subseteq mathbb Z_p^ntimes n$. Note that $T$ can be identified with the nonzero elements of $GF(p^n)$. Suppose $S$ is the subset of $T$ consisting of matrices possessing the property that every leading principal minor is nonzero. A couple of questions: (1) What is the cardinality of $S$? and (2) Does $S$ have some special meaning as a subset of $GF(p^n)$ or otherwise? Perhaps (1) depends on $f(x)$.



      Thanks for your thoughts!







      linear-algebra matrices polynomials finite-fields






      share|cite|improve this question







      New contributor




      Murphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











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      asked Mar 13 at 18:11









      MurphyMurphy

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