Determine all real numbers that can be eigenvalues of operator A.












0












$begingroup$


Problem:



Let $A$ be some linear operator such that:
$$ (A^{2006}-I)^{2006}-I=0. $$
Determine all real numbers that can be eigenvalues of operator $A$.



Question: How to solve this problem?



My attempt:



$(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$



$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$



Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?



Thank you for any help.






linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Problem:



    Let $A$ be some linear operator such that:
    $$ (A^{2006}-I)^{2006}-I=0. $$
    Determine all real numbers that can be eigenvalues of operator $A$.



    Question: How to solve this problem?



    My attempt:



    $(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$



    $A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$



    Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?



    Thank you for any help.






    linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Problem:



      Let $A$ be some linear operator such that:
      $$ (A^{2006}-I)^{2006}-I=0. $$
      Determine all real numbers that can be eigenvalues of operator $A$.



      Question: How to solve this problem?



      My attempt:



      $(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$



      $A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$



      Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?



      Thank you for any help.






      linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization







      share|cite|improve this question









      $endgroup$




      Problem:



      Let $A$ be some linear operator such that:
      $$ (A^{2006}-I)^{2006}-I=0. $$
      Determine all real numbers that can be eigenvalues of operator $A$.



      Question: How to solve this problem?



      My attempt:



      $(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$



      $A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$



      Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?



      Thank you for any help.






      linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization




      linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 20:55









      ThomThom

      322110




      322110






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          $textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
            $endgroup$
            – Thom
            Dec 16 '18 at 21:23






          • 1




            $begingroup$
            Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
            $endgroup$
            – Song
            Dec 16 '18 at 21:45



















          1












          $begingroup$

          If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            $textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
              $endgroup$
              – Thom
              Dec 16 '18 at 21:23






            • 1




              $begingroup$
              Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
              $endgroup$
              – Song
              Dec 16 '18 at 21:45
















            2












            $begingroup$

            $textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
              $endgroup$
              – Thom
              Dec 16 '18 at 21:23






            • 1




              $begingroup$
              Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
              $endgroup$
              – Song
              Dec 16 '18 at 21:45














            2












            2








            2





            $begingroup$

            $textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)






            share|cite|improve this answer









            $endgroup$



            $textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 21:05









            SongSong

            13.8k633




            13.8k633












            • $begingroup$
              Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
              $endgroup$
              – Thom
              Dec 16 '18 at 21:23






            • 1




              $begingroup$
              Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
              $endgroup$
              – Song
              Dec 16 '18 at 21:45


















            • $begingroup$
              Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
              $endgroup$
              – Thom
              Dec 16 '18 at 21:23






            • 1




              $begingroup$
              Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
              $endgroup$
              – Song
              Dec 16 '18 at 21:45
















            $begingroup$
            Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
            $endgroup$
            – Thom
            Dec 16 '18 at 21:23




            $begingroup$
            Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
            $endgroup$
            – Thom
            Dec 16 '18 at 21:23




            1




            1




            $begingroup$
            Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
            $endgroup$
            – Song
            Dec 16 '18 at 21:45




            $begingroup$
            Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
            $endgroup$
            – Song
            Dec 16 '18 at 21:45











            1












            $begingroup$

            If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$






                share|cite|improve this answer









                $endgroup$



                If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 21:08









                Ross MillikanRoss Millikan

                297k23198371




                297k23198371






























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