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If given $A^n = alpha I$ prove $A$ is diagonizable



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraMinimal Polynomial of a scalar multiple of a MatrixProperties of a matrix with the minimal polynomial $m_A(t) = t^3+2t^2+t+1$?What can we conclude from the equality $m_A(x) = m_B(x)$?If $f_A(x) ne m_A(x)$ and $A^3=I$ then $A=I$?Minimal polynomial for an invertible matrix and its determinantLet $A in mathbbM_n(F)$. Every polynomial$(p)$ of degree n such that $p(A)=0$ then how $p$ can be expressed?How do you show that if a block matrix is diagonalizable, then its diagonal entries are diagonalizable?Calculate the determinant of $A-2A^-1$ given the characteristic and minimal polynomials of $A$Prove that $P^-1v$ is an eigenvector of $B$ corresponding to the eigenvalue $lambda$.Condition for linear transformation being invertible










1












$begingroup$



Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.




I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.



I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.



How can I prove that $A$ is diagonizable?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
    $endgroup$
    – mihaild
    yesterday






  • 1




    $begingroup$
    I haven't learned the Jordan form yet, sorry
    $endgroup$
    – Guysudai1
    yesterday















1












$begingroup$



Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.




I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.



I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.



How can I prove that $A$ is diagonizable?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
    $endgroup$
    – mihaild
    yesterday






  • 1




    $begingroup$
    I haven't learned the Jordan form yet, sorry
    $endgroup$
    – Guysudai1
    yesterday













1












1








1





$begingroup$



Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.




I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.



I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.



How can I prove that $A$ is diagonizable?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$





Given matrix $ A inM_n(mathbbC), 0nealphainmathbbC$ such that $A^n = alpha I$ prove that $A$ is diagonizable.




I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.



I've also found that $|A| = alpha$ even though that didn't really give me a good starting place.



How can I prove that $A$ is diagonizable?






linear-algebra matrices diagonalization




linear-algebra matrices diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









YuiTo Cheng

2,4064937




2,4064937










asked yesterday









Guysudai1Guysudai1

296111




296111







  • 1




    $begingroup$
    Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
    $endgroup$
    – mihaild
    yesterday






  • 1




    $begingroup$
    I haven't learned the Jordan form yet, sorry
    $endgroup$
    – Guysudai1
    yesterday












  • 1




    $begingroup$
    Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
    $endgroup$
    – mihaild
    yesterday






  • 1




    $begingroup$
    I haven't learned the Jordan form yet, sorry
    $endgroup$
    – Guysudai1
    yesterday







1




1




$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday




$begingroup$
Are you allowed to use Jordan normal form? If yes - just note that all eigenvalues are non-zero, and power of Jordan block with non-zero value and size greater then $1$ is non-diagonizable.
$endgroup$
– mihaild
yesterday




1




1




$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday




$begingroup$
I haven't learned the Jordan form yet, sorry
$endgroup$
– Guysudai1
yesterday










3 Answers
3






active

oldest

votes


















5












$begingroup$

Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.



Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    (+1) Your answer is better than mine.
    $endgroup$
    – José Carlos Santos
    yesterday


















4












$begingroup$

If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...






    share|cite|improve this answer









    $endgroup$













      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.



      Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.






      share|cite|improve this answer









      $endgroup$








      • 2




        $begingroup$
        (+1) Your answer is better than mine.
        $endgroup$
        – José Carlos Santos
        yesterday















      5












      $begingroup$

      Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.



      Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.






      share|cite|improve this answer









      $endgroup$








      • 2




        $begingroup$
        (+1) Your answer is better than mine.
        $endgroup$
        – José Carlos Santos
        yesterday













      5












      5








      5





      $begingroup$

      Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.



      Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.






      share|cite|improve this answer









      $endgroup$



      Small remark: Over $mathbbC$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.



      Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered yesterday









      DirkDirk

      4,658219




      4,658219







      • 2




        $begingroup$
        (+1) Your answer is better than mine.
        $endgroup$
        – José Carlos Santos
        yesterday












      • 2




        $begingroup$
        (+1) Your answer is better than mine.
        $endgroup$
        – José Carlos Santos
        yesterday







      2




      2




      $begingroup$
      (+1) Your answer is better than mine.
      $endgroup$
      – José Carlos Santos
      yesterday




      $begingroup$
      (+1) Your answer is better than mine.
      $endgroup$
      – José Carlos Santos
      yesterday











      4












      $begingroup$

      If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.






      share|cite|improve this answer











      $endgroup$

















        4












        $begingroup$

        If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.






        share|cite|improve this answer











        $endgroup$















          4












          4








          4





          $begingroup$

          If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.






          share|cite|improve this answer











          $endgroup$



          If $A^n=alphaoperatornameId$, with $alphaneq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$beginbmatrixlambda&1&0&0&ldots&0\0&lambda&1&0&ldots&0\0&0&lambda&1&ldots&0\vdots&vdots&vdots&vdots&ddots&vdots\0&0&0&0&cdots&lambdaendbmatrix,$$with $lambdaneq0$. But no power of such a matrix is diagonlizable.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          José Carlos SantosJosé Carlos Santos

          174k23133243




          174k23133243





















              2












              $begingroup$

              Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...






                  share|cite|improve this answer









                  $endgroup$



                  Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                  35.5k42972




                  35.5k42972



























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