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prove that $A$ is diagonalizable if $A^{3}-3A^{2}-A+3I_{n} = 0$


Hermitian Matrices are DiagonalizableA question about diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove that A … is not diagonalizableProve a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix













2












$begingroup$


We have :



$A^{3}-3A^{2}-A+3I_{n} = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    6 hours ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    6 hours ago










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    2 hours ago
















2












$begingroup$


We have :



$A^{3}-3A^{2}-A+3I_{n} = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    6 hours ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    6 hours ago










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    2 hours ago














2












2








2





$begingroup$


We have :



$A^{3}-3A^{2}-A+3I_{n} = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




We have :



$A^{3}-3A^{2}-A+3I_{n} = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization




linear-algebra matrices eigenvalues-eigenvectors diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









user21820

40k544161




40k544161










asked 6 hours ago









JoshuaKJoshuaK

264




264








  • 2




    $begingroup$
    Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    6 hours ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    6 hours ago










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    2 hours ago














  • 2




    $begingroup$
    Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    6 hours ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    6 hours ago










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    2 hours ago








2




2




$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago




$begingroup$
Note that if $A=operatorname{Id}_n$, then $A^3-3A^2-A+3operatorname{Id}_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatorname{Id}_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
6 hours ago












$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago




$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
6 hours ago












$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago




$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
2 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    4 hours ago



















2












$begingroup$

Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
      $endgroup$
      – JoshuaK
      6 hours ago












    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      4 hours ago
















    3












    $begingroup$

    The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      4 hours ago














    3












    3








    3





    $begingroup$

    The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.






    share|cite|improve this answer









    $endgroup$



    The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbb{R}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 6 hours ago









    TheSilverDoeTheSilverDoe

    5,324215




    5,324215








    • 3




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      4 hours ago














    • 3




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      4 hours ago








    3




    3




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    4 hours ago




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    4 hours ago











    2












    $begingroup$

    Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






        share|cite|improve this answer









        $endgroup$



        Solving a simpler example, $A-cI_n=0$, it's clear that $a_{(i,i)}=c$ because $a_{(i,j)} - cI_{(i,j)} = 0$ for all $i,j in {1,dots,n}$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        EricEric

        513




        513























            1












            $begingroup$

            We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
              $endgroup$
              – JoshuaK
              6 hours ago
















            1












            $begingroup$

            We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
              $endgroup$
              – JoshuaK
              6 hours ago














            1












            1








            1





            $begingroup$

            We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






            share|cite|improve this answer









            $endgroup$



            We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            GSoferGSofer

            8631313




            8631313












            • $begingroup$
              Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
              $endgroup$
              – JoshuaK
              6 hours ago


















            • $begingroup$
              Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
              $endgroup$
              – JoshuaK
              6 hours ago
















            $begingroup$
            Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
            $endgroup$
            – JoshuaK
            6 hours ago




            $begingroup$
            Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
            $endgroup$
            – JoshuaK
            6 hours ago


















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