Find a determinant for a linear map $Z in M_{7×7}(mathbb R)$ if you know that $Z^{2}-8Z^{-1}$ is a zero...












1












$begingroup$


I have a trouble with this task because I think that I need clever way to do this easy and fast.

However I don't have any idea to don't count it. My only idea is to firstly calculate $Z^{2}$ but then I will have a lot of parameters because I don't have knowledge about $Z$ and I think this idea is impossible to make.

How do it in intelligent way?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
    $endgroup$
    – Will M.
    Jan 15 at 17:24


















1












$begingroup$


I have a trouble with this task because I think that I need clever way to do this easy and fast.

However I don't have any idea to don't count it. My only idea is to firstly calculate $Z^{2}$ but then I will have a lot of parameters because I don't have knowledge about $Z$ and I think this idea is impossible to make.

How do it in intelligent way?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
    $endgroup$
    – Will M.
    Jan 15 at 17:24
















1












1








1





$begingroup$


I have a trouble with this task because I think that I need clever way to do this easy and fast.

However I don't have any idea to don't count it. My only idea is to firstly calculate $Z^{2}$ but then I will have a lot of parameters because I don't have knowledge about $Z$ and I think this idea is impossible to make.

How do it in intelligent way?






linear-algebra







share|cite|improve this question











$endgroup$




I have a trouble with this task because I think that I need clever way to do this easy and fast.

However I don't have any idea to don't count it. My only idea is to firstly calculate $Z^{2}$ but then I will have a lot of parameters because I don't have knowledge about $Z$ and I think this idea is impossible to make.

How do it in intelligent way?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 17:06







MP3129

















asked Jan 15 at 16:52









MP3129MP3129

900211




900211








  • 1




    $begingroup$
    $Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
    $endgroup$
    – Will M.
    Jan 15 at 17:24
















  • 1




    $begingroup$
    $Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
    $endgroup$
    – Will M.
    Jan 15 at 17:24










1




1




$begingroup$
$Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
$endgroup$
– Will M.
Jan 15 at 17:24






$begingroup$
$Z^2=8Z^{-1}$ hence the determinant satisfies the equation $x^2 = 8^7/x.$
$endgroup$
– Will M.
Jan 15 at 17:24












2 Answers
2






active

oldest

votes


















2












$begingroup$

Then $Z^3=8I$, so $(det Z)^3 = 8^7$. Hence $det Z = 2^7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How you get to the power of $7$?
    $endgroup$
    – MP3129
    Jan 15 at 22:28








  • 1




    $begingroup$
    In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
    $endgroup$
    – steven gregory
    Jan 15 at 22:37



















0












$begingroup$

Then $Z^3=8I$, so $Z$ is diagonalisable, and its eigenvalues are in the set
${2,2omega,2omega^2}$ where $omega=exp(2pi i/3)$. The determinant of
$Z$ is a product of seven of these, and is real....






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












    $begingroup$

    Then $Z^3=8I$, so $(det Z)^3 = 8^7$. Hence $det Z = 2^7$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How you get to the power of $7$?
      $endgroup$
      – MP3129
      Jan 15 at 22:28








    • 1




      $begingroup$
      In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
      $endgroup$
      – steven gregory
      Jan 15 at 22:37
















    2












    $begingroup$

    Then $Z^3=8I$, so $(det Z)^3 = 8^7$. Hence $det Z = 2^7$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How you get to the power of $7$?
      $endgroup$
      – MP3129
      Jan 15 at 22:28








    • 1




      $begingroup$
      In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
      $endgroup$
      – steven gregory
      Jan 15 at 22:37














    2












    2








    2





    $begingroup$

    Then $Z^3=8I$, so $(det Z)^3 = 8^7$. Hence $det Z = 2^7$.






    share|cite|improve this answer









    $endgroup$



    Then $Z^3=8I$, so $(det Z)^3 = 8^7$. Hence $det Z = 2^7$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 15 at 17:13









    steven gregorysteven gregory

    18.5k32459




    18.5k32459












    • $begingroup$
      How you get to the power of $7$?
      $endgroup$
      – MP3129
      Jan 15 at 22:28








    • 1




      $begingroup$
      In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
      $endgroup$
      – steven gregory
      Jan 15 at 22:37


















    • $begingroup$
      How you get to the power of $7$?
      $endgroup$
      – MP3129
      Jan 15 at 22:28








    • 1




      $begingroup$
      In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
      $endgroup$
      – steven gregory
      Jan 15 at 22:37
















    $begingroup$
    How you get to the power of $7$?
    $endgroup$
    – MP3129
    Jan 15 at 22:28






    $begingroup$
    How you get to the power of $7$?
    $endgroup$
    – MP3129
    Jan 15 at 22:28






    1




    1




    $begingroup$
    In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
    $endgroup$
    – steven gregory
    Jan 15 at 22:37




    $begingroup$
    In $M_{7 times 7}$, $8I$ is a matrix with all off-diagonal elements equal to $0$ and all (seven) diagonal elements equal to $8$. Its determinant is the product of the diagonal elements.
    $endgroup$
    – steven gregory
    Jan 15 at 22:37











    0












    $begingroup$

    Then $Z^3=8I$, so $Z$ is diagonalisable, and its eigenvalues are in the set
    ${2,2omega,2omega^2}$ where $omega=exp(2pi i/3)$. The determinant of
    $Z$ is a product of seven of these, and is real....






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Then $Z^3=8I$, so $Z$ is diagonalisable, and its eigenvalues are in the set
      ${2,2omega,2omega^2}$ where $omega=exp(2pi i/3)$. The determinant of
      $Z$ is a product of seven of these, and is real....






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Then $Z^3=8I$, so $Z$ is diagonalisable, and its eigenvalues are in the set
        ${2,2omega,2omega^2}$ where $omega=exp(2pi i/3)$. The determinant of
        $Z$ is a product of seven of these, and is real....






        share|cite|improve this answer









        $endgroup$



        Then $Z^3=8I$, so $Z$ is diagonalisable, and its eigenvalues are in the set
        ${2,2omega,2omega^2}$ where $omega=exp(2pi i/3)$. The determinant of
        $Z$ is a product of seven of these, and is real....







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 17:01









        Lord Shark the UnknownLord Shark the Unknown

        109k1163136




        109k1163136






























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