Let $Min M(n,Bbb{R})$ such that $M^3=I$ and $Mvne v$ for any non-zero vector $v$. Prove that $n$ is even.












2












$begingroup$


I have tried the problem in the following manner but I haven't got any solution.



Let $lambda$ be an eigenvalue of $M$, then $exists vne 0$ such that $Mv=lambda vimplies M^3 v=lambda^3vimplies Iv=lambda^3v$
$implieslambda^3$ is an eigenvalue of the $ntimes n$ identity matrix
$Iimplies |I-lambda^3I|=0implies(1-lambda^3)^n=0$ $implies(lambda^2+lambda+1)^n=0$ since $lambdane 1$, given that $Mvne v$.



Now, from this I can conclude nothing?

Can anybody derive a solution for this problem?






linear-algebra matrices eigenvalues-eigenvectors







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$endgroup$












  • $begingroup$
    What can you say about the characteristic polynomial of $M$?
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 12:29










  • $begingroup$
    @LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:31
















2












$begingroup$


I have tried the problem in the following manner but I haven't got any solution.



Let $lambda$ be an eigenvalue of $M$, then $exists vne 0$ such that $Mv=lambda vimplies M^3 v=lambda^3vimplies Iv=lambda^3v$
$implieslambda^3$ is an eigenvalue of the $ntimes n$ identity matrix
$Iimplies |I-lambda^3I|=0implies(1-lambda^3)^n=0$ $implies(lambda^2+lambda+1)^n=0$ since $lambdane 1$, given that $Mvne v$.



Now, from this I can conclude nothing?

Can anybody derive a solution for this problem?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    What can you say about the characteristic polynomial of $M$?
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 12:29










  • $begingroup$
    @LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:31














2












2








2





$begingroup$


I have tried the problem in the following manner but I haven't got any solution.



Let $lambda$ be an eigenvalue of $M$, then $exists vne 0$ such that $Mv=lambda vimplies M^3 v=lambda^3vimplies Iv=lambda^3v$
$implieslambda^3$ is an eigenvalue of the $ntimes n$ identity matrix
$Iimplies |I-lambda^3I|=0implies(1-lambda^3)^n=0$ $implies(lambda^2+lambda+1)^n=0$ since $lambdane 1$, given that $Mvne v$.



Now, from this I can conclude nothing?

Can anybody derive a solution for this problem?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I have tried the problem in the following manner but I haven't got any solution.



Let $lambda$ be an eigenvalue of $M$, then $exists vne 0$ such that $Mv=lambda vimplies M^3 v=lambda^3vimplies Iv=lambda^3v$
$implieslambda^3$ is an eigenvalue of the $ntimes n$ identity matrix
$Iimplies |I-lambda^3I|=0implies(1-lambda^3)^n=0$ $implies(lambda^2+lambda+1)^n=0$ since $lambdane 1$, given that $Mvne v$.



Now, from this I can conclude nothing?

Can anybody derive a solution for this problem?






linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 18:20









Hanno

2,106425




2,106425










asked Nov 25 '18 at 12:25









Biswarup SahaBiswarup Saha

557110




557110












  • $begingroup$
    What can you say about the characteristic polynomial of $M$?
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 12:29










  • $begingroup$
    @LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:31


















  • $begingroup$
    What can you say about the characteristic polynomial of $M$?
    $endgroup$
    – Lord Shark the Unknown
    Nov 25 '18 at 12:29










  • $begingroup$
    @LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:31
















$begingroup$
What can you say about the characteristic polynomial of $M$?
$endgroup$
– Lord Shark the Unknown
Nov 25 '18 at 12:29




$begingroup$
What can you say about the characteristic polynomial of $M$?
$endgroup$
– Lord Shark the Unknown
Nov 25 '18 at 12:29












$begingroup$
@LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
$endgroup$
– Biswarup Saha
Nov 25 '18 at 12:31




$begingroup$
@LordSharktheUnknown I just can think if $n$ is odd then it must posses at least one real eigenvalue.
$endgroup$
– Biswarup Saha
Nov 25 '18 at 12:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

If $n$ is odd, then $M$ has some real eigenvalue $lambda$. Since $M^3=operatorname{Id}_3$, $lambda^3=1$, which is equivalent to the assertion that $lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=lambda v=v$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:41










  • $begingroup$
    I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
    $endgroup$
    – José Carlos Santos
    Nov 25 '18 at 12:43










  • $begingroup$
    Jose Carlos Santos, Ok I got it.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 13:09



















0












$begingroup$

By the hypotheses,





  • $M$ is invertible with inverse $M^{-1}=M^2$,


  • $M-I,$ is invertible as $1$ is not an eigenvalue.


This yields
$$0:=:M^3-I :=: (M-I)(M^2+M+I):: implies M^{-1}+M+I = 0$$
thus eigenvalues of $M$ must satisfy
$$frac 1lambda + lambda:=: -1quadLongleftrightarrowquadlambda:=: frac{-1pmsqrt 3i}2 :=: expleft(pm, ifrac{2pi} 3right),.
$$
Hence $M,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n,$ cannot be odd.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    If $n$ is odd, then $M$ has some real eigenvalue $lambda$. Since $M^3=operatorname{Id}_3$, $lambda^3=1$, which is equivalent to the assertion that $lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=lambda v=v$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 12:41










    • $begingroup$
      I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
      $endgroup$
      – José Carlos Santos
      Nov 25 '18 at 12:43










    • $begingroup$
      Jose Carlos Santos, Ok I got it.
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 13:09
















    1












    $begingroup$

    If $n$ is odd, then $M$ has some real eigenvalue $lambda$. Since $M^3=operatorname{Id}_3$, $lambda^3=1$, which is equivalent to the assertion that $lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=lambda v=v$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 12:41










    • $begingroup$
      I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
      $endgroup$
      – José Carlos Santos
      Nov 25 '18 at 12:43










    • $begingroup$
      Jose Carlos Santos, Ok I got it.
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 13:09














    1












    1








    1





    $begingroup$

    If $n$ is odd, then $M$ has some real eigenvalue $lambda$. Since $M^3=operatorname{Id}_3$, $lambda^3=1$, which is equivalent to the assertion that $lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=lambda v=v$.






    share|cite|improve this answer









    $endgroup$



    If $n$ is odd, then $M$ has some real eigenvalue $lambda$. Since $M^3=operatorname{Id}_3$, $lambda^3=1$, which is equivalent to the assertion that $lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=lambda v=v$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 '18 at 12:33









    José Carlos SantosJosé Carlos Santos

    154k22124227




    154k22124227












    • $begingroup$
      I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 12:41










    • $begingroup$
      I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
      $endgroup$
      – José Carlos Santos
      Nov 25 '18 at 12:43










    • $begingroup$
      Jose Carlos Santos, Ok I got it.
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 13:09


















    • $begingroup$
      I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 12:41










    • $begingroup$
      I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
      $endgroup$
      – José Carlos Santos
      Nov 25 '18 at 12:43










    • $begingroup$
      Jose Carlos Santos, Ok I got it.
      $endgroup$
      – Biswarup Saha
      Nov 25 '18 at 13:09
















    $begingroup$
    I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:41




    $begingroup$
    I think I have got it. $(lambda^2+lambda+1)^n=0implies$ all values of $lambda$ are $omega, omega^2$. But does it give all eigen values? May be YES? Am I right?
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 12:41












    $begingroup$
    I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
    $endgroup$
    – José Carlos Santos
    Nov 25 '18 at 12:43




    $begingroup$
    I don't know why you felt the need to introduce $lambda^2+lambda+1$ or $omega$ here. Since $lambda^3=1$ and $lambdainmathbb R$, $lambda=1$ (every real number has one and only one cube root).
    $endgroup$
    – José Carlos Santos
    Nov 25 '18 at 12:43












    $begingroup$
    Jose Carlos Santos, Ok I got it.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 13:09




    $begingroup$
    Jose Carlos Santos, Ok I got it.
    $endgroup$
    – Biswarup Saha
    Nov 25 '18 at 13:09











    0












    $begingroup$

    By the hypotheses,





    • $M$ is invertible with inverse $M^{-1}=M^2$,


    • $M-I,$ is invertible as $1$ is not an eigenvalue.


    This yields
    $$0:=:M^3-I :=: (M-I)(M^2+M+I):: implies M^{-1}+M+I = 0$$
    thus eigenvalues of $M$ must satisfy
    $$frac 1lambda + lambda:=: -1quadLongleftrightarrowquadlambda:=: frac{-1pmsqrt 3i}2 :=: expleft(pm, ifrac{2pi} 3right),.
    $$
    Hence $M,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n,$ cannot be odd.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By the hypotheses,





      • $M$ is invertible with inverse $M^{-1}=M^2$,


      • $M-I,$ is invertible as $1$ is not an eigenvalue.


      This yields
      $$0:=:M^3-I :=: (M-I)(M^2+M+I):: implies M^{-1}+M+I = 0$$
      thus eigenvalues of $M$ must satisfy
      $$frac 1lambda + lambda:=: -1quadLongleftrightarrowquadlambda:=: frac{-1pmsqrt 3i}2 :=: expleft(pm, ifrac{2pi} 3right),.
      $$
      Hence $M,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n,$ cannot be odd.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By the hypotheses,





        • $M$ is invertible with inverse $M^{-1}=M^2$,


        • $M-I,$ is invertible as $1$ is not an eigenvalue.


        This yields
        $$0:=:M^3-I :=: (M-I)(M^2+M+I):: implies M^{-1}+M+I = 0$$
        thus eigenvalues of $M$ must satisfy
        $$frac 1lambda + lambda:=: -1quadLongleftrightarrowquadlambda:=: frac{-1pmsqrt 3i}2 :=: expleft(pm, ifrac{2pi} 3right),.
        $$
        Hence $M,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n,$ cannot be odd.






        share|cite|improve this answer









        $endgroup$



        By the hypotheses,





        • $M$ is invertible with inverse $M^{-1}=M^2$,


        • $M-I,$ is invertible as $1$ is not an eigenvalue.


        This yields
        $$0:=:M^3-I :=: (M-I)(M^2+M+I):: implies M^{-1}+M+I = 0$$
        thus eigenvalues of $M$ must satisfy
        $$frac 1lambda + lambda:=: -1quadLongleftrightarrowquadlambda:=: frac{-1pmsqrt 3i}2 :=: expleft(pm, ifrac{2pi} 3right),.
        $$
        Hence $M,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n,$ cannot be odd.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 19:25









        HannoHanno

        2,106425




        2,106425






























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