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prove that the matrix A is diagonalizable

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prove that the matrix A is diagonalizable


Block Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove 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?Is the Matrix Diagonalizable if $A^2=4I$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











$endgroup$







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
    55 mins ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    53 mins 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











$endgroup$







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
    55 mins ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    53 mins 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 53 mins ago







JoshuaK

















asked 58 mins ago









JoshuaKJoshuaK

254




254







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
    55 mins ago










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












  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
    55 mins ago










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







1




1




$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
55 mins ago




$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ 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
55 mins ago












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




$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
53 mins 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 $mathbbR$.






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









    $endgroup$




















      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
        32 mins 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 $mathbbR$.






      share|cite|improve this answer









      $endgroup$

















        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 $mathbbR$.






        share|cite|improve this answer









        $endgroup$















          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 $mathbbR$.






          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 $mathbbR$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 52 mins ago









          TheSilverDoeTheSilverDoe

          5,324215




          5,324215





















              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 47 mins 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
                        32 mins 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
                        32 mins 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 50 mins 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
                        32 mins ago
















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















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




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

















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