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












7












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    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
    Feb 5 at 21:41








  • 3




    $begingroup$
    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
    Feb 6 at 0:23


















7












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    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
    Feb 5 at 21:41








  • 3




    $begingroup$
    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
    Feb 6 at 0:23
















7












7








7





$begingroup$


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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 5 at 22:50









J. W. Tanner

3,9221320




3,9221320










asked Feb 5 at 20:30









IdoIdo

1025




1025








  • 3




    $begingroup$
    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
    Feb 5 at 21:41








  • 3




    $begingroup$
    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
    Feb 6 at 0:23
















  • 3




    $begingroup$
    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
    Feb 5 at 21:41








  • 3




    $begingroup$
    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
    Feb 6 at 0:23










3




3




$begingroup$
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
Feb 5 at 21:41






$begingroup$
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
Feb 5 at 21:41






3




3




$begingroup$
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
Feb 6 at 0:23






$begingroup$
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
Feb 6 at 0:23












5 Answers
5






active

oldest

votes


















16












$begingroup$

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
    Feb 5 at 20:40










  • $begingroup$
    Are all counterexamples of this form?
    $endgroup$
    – PyRulez
    Feb 6 at 0:26






  • 2




    $begingroup$
    @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
    Feb 6 at 0:39





















9












$begingroup$

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






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    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









    $endgroup$









    • 1




      $begingroup$
      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
      Feb 5 at 20:43





















    5












    $begingroup$

    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









    $endgroup$





















      5












      $begingroup$

      No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.



      Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.



      For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.



      However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.






      share|cite|improve this answer









      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        16












        $begingroup$

        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
          Feb 5 at 20:40










        • $begingroup$
          Are all counterexamples of this form?
          $endgroup$
          – PyRulez
          Feb 6 at 0:26






        • 2




          $begingroup$
          @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
          Feb 6 at 0:39


















        16












        $begingroup$

        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
          Feb 5 at 20:40










        • $begingroup$
          Are all counterexamples of this form?
          $endgroup$
          – PyRulez
          Feb 6 at 0:26






        • 2




          $begingroup$
          @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
          Feb 6 at 0:39
















        16












        16








        16





        $begingroup$

        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 Feb 5 at 20:35









        MarkMark

        10.4k1622




        10.4k1622












        • $begingroup$
          True, thank you
          $endgroup$
          – Ido
          Feb 5 at 20:40










        • $begingroup$
          Are all counterexamples of this form?
          $endgroup$
          – PyRulez
          Feb 6 at 0:26






        • 2




          $begingroup$
          @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
          Feb 6 at 0:39




















        • $begingroup$
          True, thank you
          $endgroup$
          – Ido
          Feb 5 at 20:40










        • $begingroup$
          Are all counterexamples of this form?
          $endgroup$
          – PyRulez
          Feb 6 at 0:26






        • 2




          $begingroup$
          @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
          Feb 6 at 0:39


















        $begingroup$
        True, thank you
        $endgroup$
        – Ido
        Feb 5 at 20:40




        $begingroup$
        True, thank you
        $endgroup$
        – Ido
        Feb 5 at 20:40












        $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        Feb 6 at 0:26




        $begingroup$
        Are all counterexamples of this form?
        $endgroup$
        – PyRulez
        Feb 6 at 0:26




        2




        2




        $begingroup$
        @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
        Feb 6 at 0:39






        $begingroup$
        @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
        Feb 6 at 0:39













        9












        $begingroup$

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






        share|cite|improve this answer









        $endgroup$


















          9












          $begingroup$

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






          share|cite|improve this answer









          $endgroup$
















            9












            9








            9





            $begingroup$

            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 Feb 5 at 20:34









            MicappsMicapps

            1,15939




            1,15939























                5












                $begingroup$

                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









                $endgroup$









                • 1




                  $begingroup$
                  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
                  Feb 5 at 20:43


















                5












                $begingroup$

                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









                $endgroup$









                • 1




                  $begingroup$
                  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
                  Feb 5 at 20:43
















                5












                5








                5





                $begingroup$

                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 Feb 5 at 20:33









                Robert ZRobert Z

                101k1070143




                101k1070143








                • 1




                  $begingroup$
                  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
                  Feb 5 at 20:43
















                • 1




                  $begingroup$
                  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
                  Feb 5 at 20:43










                1




                1




                $begingroup$
                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
                Feb 5 at 20:43






                $begingroup$
                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
                Feb 5 at 20:43













                5












                $begingroup$

                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









                $endgroup$


















                  5












                  $begingroup$

                  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









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    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 Feb 5 at 20:50









                    Robert IsraelRobert Israel

                    329k23218472




                    329k23218472























                        5












                        $begingroup$

                        No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.



                        Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.



                        For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.



                        However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.






                        share|cite|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.



                          Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.



                          For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.



                          However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.






                          share|cite|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.



                            Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.



                            For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.



                            However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.






                            share|cite|improve this answer









                            $endgroup$



                            No. In general if a matrix is a root of a polynomial, that polynomial is a multiple of the minimal polynomial of that matrix.



                            Even if you only consider n-degree polynomials for an n×n matrix, this still can fail.



                            For example, if A=0, then the characteristic polynomial is $x^n$, but A is the root of any polynomial $xg(x)$.



                            However, all is not lost. If your minimal polynomial equals your characteristic polynomial (as with companion matrices or in the case you have no repeated roots), then the characteristic polynomial (= minimal polynomial) will be the unique monic n-degree polynomial with root A.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 6 at 4:47









                            Bill CookBill Cook

                            23.2k4869




                            23.2k4869






























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                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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                                File:DeusFollowingSea.jpg

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                                File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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