Does the Cayley–Hamilton theorem work in the opposite direction?












5












$begingroup$


The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.



But does it work in the opposite direction?



If for example for a certain matrix $A$ we know that



$ A^2-6A+9I=0, $



does that mean that the characteristic equation of $A$ is



$
lambda^2-6lambda+9=0
$
?






linear-algebra cayley-hamilton







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  • $begingroup$
    Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
    $endgroup$
    – ShreevatsaR
    5 hours ago












  • $begingroup$
    The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
    $endgroup$
    – JiK
    2 hours ago


















5












$begingroup$


The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.



But does it work in the opposite direction?



If for example for a certain matrix $A$ we know that



$ A^2-6A+9I=0, $



does that mean that the characteristic equation of $A$ is



$
lambda^2-6lambda+9=0
$
?






linear-algebra cayley-hamilton







share|cite|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
    $endgroup$
    – ShreevatsaR
    5 hours ago












  • $begingroup$
    The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
    $endgroup$
    – JiK
    2 hours ago
















5












5








5





$begingroup$


The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.



But does it work in the opposite direction?



If for example for a certain matrix $A$ we know that



$ A^2-6A+9I=0, $



does that mean that the characteristic equation of $A$ is



$
lambda^2-6lambda+9=0
$
?






linear-algebra cayley-hamilton







share|cite|improve this question









New contributor




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







$endgroup$




The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.



But does it work in the opposite direction?



If for example for a certain matrix $A$ we know that



$ A^2-6A+9I=0, $



does that mean that the characteristic equation of $A$ is



$
lambda^2-6lambda+9=0
$
?






linear-algebra cayley-hamilton




linear-algebra cayley-hamilton






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




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









share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









J. W. Tanner

1,470114




1,470114






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asked 6 hours ago









IdoIdo

693




693




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Ido is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






Ido is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
    $endgroup$
    – ShreevatsaR
    5 hours ago












  • $begingroup$
    The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
    $endgroup$
    – JiK
    2 hours ago




















  • $begingroup$
    Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
    $endgroup$
    – ShreevatsaR
    5 hours ago












  • $begingroup$
    The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
    $endgroup$
    – JiK
    2 hours ago


















$begingroup$
Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
$endgroup$
– ShreevatsaR
5 hours ago






$begingroup$
Yes there's an “opposite direction” that works, namely that the minimal polynomial (which divides the characteristic polynomial and often equals it) of $A$ divides (rather than necessarily equals) $lambda^2 - 6lambda + 9 = 0$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
$endgroup$
– ShreevatsaR
5 hours ago














$begingroup$
The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
$endgroup$
– JiK
2 hours ago






$begingroup$
The answers would be a lot more instructive if the question had a condition that the degree of the polynomial is the same as the size of the matrix. Too late now, I'm afraid.
$endgroup$
– JiK
2 hours ago












4 Answers
4






active

oldest

votes


















7












$begingroup$

Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    True, thank you
    $endgroup$
    – Ido
    6 hours ago










  • $begingroup$
    Are all counterexamples of this form?
    $endgroup$
    – PyRulez
    2 hours ago






  • 1




    $begingroup$
    @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
    $endgroup$
    – Mark
    2 hours ago





















3












$begingroup$

No, the $ntimes n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $nnot=2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
    $endgroup$
    – Robert Z
    6 hours ago





















3












$begingroup$

No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    For any $n times n$ matrix $A$, the $(2n) times (2n)$ matrix $pmatrix{A & 0cr 0 & Acr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.






    share|cite|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        True, thank you
        $endgroup$
        – Ido
        6 hours ago










      • $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        2 hours ago






      • 1




        $begingroup$
        @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
        $endgroup$
        – Mark
        2 hours ago


















      7












      $begingroup$

      Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        True, thank you
        $endgroup$
        – Ido
        6 hours ago










      • $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        2 hours ago






      • 1




        $begingroup$
        @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
        $endgroup$
        – Mark
        2 hours ago
















      7












      7








      7





      $begingroup$

      Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.






      share|cite|improve this answer









      $endgroup$



      Even without counterexamples it is obvious that your statement can't be true because if $A$ is a root of the polynomial $p(x)$ then it must be the root of $p(x)q(x)$ for any polynomial $q$. So that way we would get the matrix $A$ has infinitely many characteristic polynomials.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 6 hours ago









      MarkMark

      6,998416




      6,998416












      • $begingroup$
        True, thank you
        $endgroup$
        – Ido
        6 hours ago










      • $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        2 hours ago






      • 1




        $begingroup$
        @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
        $endgroup$
        – Mark
        2 hours ago




















      • $begingroup$
        True, thank you
        $endgroup$
        – Ido
        6 hours ago










      • $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        2 hours ago






      • 1




        $begingroup$
        @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
        $endgroup$
        – Mark
        2 hours ago


















      $begingroup$
      True, thank you
      $endgroup$
      – Ido
      6 hours ago




      $begingroup$
      True, thank you
      $endgroup$
      – Ido
      6 hours ago












      $begingroup$
      Are all counterexamples of this form?
      $endgroup$
      – PyRulez
      2 hours ago




      $begingroup$
      Are all counterexamples of this form?
      $endgroup$
      – PyRulez
      2 hours ago




      1




      1




      $begingroup$
      @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
      $endgroup$
      – Mark
      2 hours ago






      $begingroup$
      @PyRulez, the set of polynomials $p$ such that $p(A)=0$ is ${mq$: q is a polynomial$}$ when $m$ is the minimal polynomial of $A$. So all such polynomials are multiples of the minimal polynomial.
      $endgroup$
      – Mark
      2 hours ago













      3












      $begingroup$

      No, the $ntimes n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $nnot=2$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
        $endgroup$
        – Robert Z
        6 hours ago


















      3












      $begingroup$

      No, the $ntimes n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $nnot=2$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
        $endgroup$
        – Robert Z
        6 hours ago
















      3












      3








      3





      $begingroup$

      No, the $ntimes n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $nnot=2$.






      share|cite|improve this answer









      $endgroup$



      No, the $ntimes n$ matrix $A=3I$ satisfies the given equation but it has a different characteristic polynomial for $nnot=2$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 6 hours ago









      Robert ZRobert Z

      96.3k1065136




      96.3k1065136












      • $begingroup$
        If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
        $endgroup$
        – Robert Z
        6 hours ago




















      • $begingroup$
        If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
        $endgroup$
        – Robert Z
        6 hours ago


















      $begingroup$
      If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
      $endgroup$
      – Robert Z
      6 hours ago






      $begingroup$
      If $P(A)=0$ then the minimal polynomial of $A$ divides $Q$. See en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra) If the characteristic poynomial has distict roots then it divides $P$.
      $endgroup$
      – Robert Z
      6 hours ago













      3












      $begingroup$

      No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.






          share|cite|improve this answer









          $endgroup$



          No. For example, $I-1=0$, but the characteristic polynomial of $I$ is $(x-1)^n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          MicappsMicapps

          89137




          89137























              2












              $begingroup$

              For any $n times n$ matrix $A$, the $(2n) times (2n)$ matrix $pmatrix{A & 0cr 0 & Acr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For any $n times n$ matrix $A$, the $(2n) times (2n)$ matrix $pmatrix{A & 0cr 0 & Acr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For any $n times n$ matrix $A$, the $(2n) times (2n)$ matrix $pmatrix{A & 0cr 0 & Acr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.






                  share|cite|improve this answer









                  $endgroup$



                  For any $n times n$ matrix $A$, the $(2n) times (2n)$ matrix $pmatrix{A & 0cr 0 & Acr}$ satisfies the characteristic polynomial of $A$, but its own characteristic polynomial is the square of that of $A$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  Robert IsraelRobert Israel

                  322k23212465




                  322k23212465






















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