Finding triples (a,b,c) so that characteristic and minimal polynomials are different












1












$begingroup$


I got to find all complex triple $(a,b,c)$ so that the following matrix has different characteristic and minimal polynomial.begin{bmatrix}2&0&0\a&2&0\b&c&-1\end{bmatrix}



Certainly the eigenvalues are 2,2,-1.
Hence the characteristic and minimal polynomials are same if they are $(x-2)^2(x+1)$. And the JCF would be
$ begin{bmatrix}2&0&0\1&2&0\0&0&-1\end{bmatrix}$.



And the characteristic and minimal polynomials would be different if the minimal polynomial would be $(x-2)(x+1)$ and the JCF would be $ begin{bmatrix}2&0&0\0&2&0\0&0&-1\end{bmatrix}$.



Now how to find all the complex triples which works in the second case i.e. the minimal polynomial is $(x+1)(x-2)$.






linear-algebra eigenvalues-eigenvectors







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  • 3




    $begingroup$
    So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:58
















1












$begingroup$


I got to find all complex triple $(a,b,c)$ so that the following matrix has different characteristic and minimal polynomial.begin{bmatrix}2&0&0\a&2&0\b&c&-1\end{bmatrix}



Certainly the eigenvalues are 2,2,-1.
Hence the characteristic and minimal polynomials are same if they are $(x-2)^2(x+1)$. And the JCF would be
$ begin{bmatrix}2&0&0\1&2&0\0&0&-1\end{bmatrix}$.



And the characteristic and minimal polynomials would be different if the minimal polynomial would be $(x-2)(x+1)$ and the JCF would be $ begin{bmatrix}2&0&0\0&2&0\0&0&-1\end{bmatrix}$.



Now how to find all the complex triples which works in the second case i.e. the minimal polynomial is $(x+1)(x-2)$.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:58














1












1








1


1



$begingroup$


I got to find all complex triple $(a,b,c)$ so that the following matrix has different characteristic and minimal polynomial.begin{bmatrix}2&0&0\a&2&0\b&c&-1\end{bmatrix}



Certainly the eigenvalues are 2,2,-1.
Hence the characteristic and minimal polynomials are same if they are $(x-2)^2(x+1)$. And the JCF would be
$ begin{bmatrix}2&0&0\1&2&0\0&0&-1\end{bmatrix}$.



And the characteristic and minimal polynomials would be different if the minimal polynomial would be $(x-2)(x+1)$ and the JCF would be $ begin{bmatrix}2&0&0\0&2&0\0&0&-1\end{bmatrix}$.



Now how to find all the complex triples which works in the second case i.e. the minimal polynomial is $(x+1)(x-2)$.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I got to find all complex triple $(a,b,c)$ so that the following matrix has different characteristic and minimal polynomial.begin{bmatrix}2&0&0\a&2&0\b&c&-1\end{bmatrix}



Certainly the eigenvalues are 2,2,-1.
Hence the characteristic and minimal polynomials are same if they are $(x-2)^2(x+1)$. And the JCF would be
$ begin{bmatrix}2&0&0\1&2&0\0&0&-1\end{bmatrix}$.



And the characteristic and minimal polynomials would be different if the minimal polynomial would be $(x-2)(x+1)$ and the JCF would be $ begin{bmatrix}2&0&0\0&2&0\0&0&-1\end{bmatrix}$.



Now how to find all the complex triples which works in the second case i.e. the minimal polynomial is $(x+1)(x-2)$.






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 4:06







ChakSayantan

















asked Nov 30 '18 at 3:56









ChakSayantanChakSayantan

1426




1426








  • 3




    $begingroup$
    So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:58














  • 3




    $begingroup$
    So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:58








3




3




$begingroup$
So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
$endgroup$
– Anurag A
Nov 30 '18 at 3:58




$begingroup$
So you want $(A-I)(A-2I)$ to be the zero transformation. This should give you some conditions on $a,b,c$.
$endgroup$
– Anurag A
Nov 30 '18 at 3:58










1 Answer
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$begingroup$

Since you edited the problem, so things will change.



Consider the matrix
$$(A+I)(A-2I)=begin{bmatrix}3&0&0\a&3&0\b&c&0end{bmatrix}begin{bmatrix}0&0&0\a&0&0\b&c&-3end{bmatrix}=begin{bmatrix}0&0&0\3a&0&0\ac&0&0\end{bmatrix}$$
With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.






share|cite|improve this answer











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    $begingroup$

    Since you edited the problem, so things will change.



    Consider the matrix
    $$(A+I)(A-2I)=begin{bmatrix}3&0&0\a&3&0\b&c&0end{bmatrix}begin{bmatrix}0&0&0\a&0&0\b&c&-3end{bmatrix}=begin{bmatrix}0&0&0\3a&0&0\ac&0&0\end{bmatrix}$$
    With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Since you edited the problem, so things will change.



      Consider the matrix
      $$(A+I)(A-2I)=begin{bmatrix}3&0&0\a&3&0\b&c&0end{bmatrix}begin{bmatrix}0&0&0\a&0&0\b&c&-3end{bmatrix}=begin{bmatrix}0&0&0\3a&0&0\ac&0&0\end{bmatrix}$$
      With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Since you edited the problem, so things will change.



        Consider the matrix
        $$(A+I)(A-2I)=begin{bmatrix}3&0&0\a&3&0\b&c&0end{bmatrix}begin{bmatrix}0&0&0\a&0&0\b&c&-3end{bmatrix}=begin{bmatrix}0&0&0\3a&0&0\ac&0&0\end{bmatrix}$$
        With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.






        share|cite|improve this answer











        $endgroup$



        Since you edited the problem, so things will change.



        Consider the matrix
        $$(A+I)(A-2I)=begin{bmatrix}3&0&0\a&3&0\b&c&0end{bmatrix}begin{bmatrix}0&0&0\a&0&0\b&c&-3end{bmatrix}=begin{bmatrix}0&0&0\3a&0&0\ac&0&0\end{bmatrix}$$
        With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 4:11

























        answered Nov 30 '18 at 4:03









        Anurag AAnurag A

        25.8k12249




        25.8k12249






























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