Eigenvectors of the following diagonal matrix












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$begingroup$


Let $x = frac{2 i pi k}{n}$ where $1 leq k leq frac{n}{2} - 1$, where $n = 2k$ even. $$ A = begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix}$$ I'm trying to find the eigenvectors of of this matrix. The eigenvalues are clearly $e^{ix}$ and $e^{-ix}$. So, let $ begin{bmatrix} a \ b end{bmatrix}$ denote the eigenvector. Then $$ begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix} begin{bmatrix} a \ b end{bmatrix} = e^{ix} begin{bmatrix} a \ b end{bmatrix}$$ So, we get that $a = 1$ from the first equation. Solving for $b$ is my problem. I get $e^{-ix}b = e^{ix}b$, so $b = e^{2ix}b$, then $b(1 - e^{2ix}) = 0$, but then I get stuck. The eigenvalues are supposedly $$begin{bmatrix} 1 \ i end{bmatrix}, begin{bmatrix} 1 \ -i end{bmatrix}$$






linear-algebra abstract-algebra eigenvalues-eigenvectors







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  • $begingroup$
    $[1, 0], [0, 1]$ always work for diagonal matrices.
    $endgroup$
    – xbh
    Dec 1 '18 at 15:01










  • $begingroup$
    $i$ appears twice and your second equation is wrong. Review what you wrote.
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:02












  • $begingroup$
    And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:07
















0












$begingroup$


Let $x = frac{2 i pi k}{n}$ where $1 leq k leq frac{n}{2} - 1$, where $n = 2k$ even. $$ A = begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix}$$ I'm trying to find the eigenvectors of of this matrix. The eigenvalues are clearly $e^{ix}$ and $e^{-ix}$. So, let $ begin{bmatrix} a \ b end{bmatrix}$ denote the eigenvector. Then $$ begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix} begin{bmatrix} a \ b end{bmatrix} = e^{ix} begin{bmatrix} a \ b end{bmatrix}$$ So, we get that $a = 1$ from the first equation. Solving for $b$ is my problem. I get $e^{-ix}b = e^{ix}b$, so $b = e^{2ix}b$, then $b(1 - e^{2ix}) = 0$, but then I get stuck. The eigenvalues are supposedly $$begin{bmatrix} 1 \ i end{bmatrix}, begin{bmatrix} 1 \ -i end{bmatrix}$$






linear-algebra abstract-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    $[1, 0], [0, 1]$ always work for diagonal matrices.
    $endgroup$
    – xbh
    Dec 1 '18 at 15:01










  • $begingroup$
    $i$ appears twice and your second equation is wrong. Review what you wrote.
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:02












  • $begingroup$
    And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:07














0












0








0





$begingroup$


Let $x = frac{2 i pi k}{n}$ where $1 leq k leq frac{n}{2} - 1$, where $n = 2k$ even. $$ A = begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix}$$ I'm trying to find the eigenvectors of of this matrix. The eigenvalues are clearly $e^{ix}$ and $e^{-ix}$. So, let $ begin{bmatrix} a \ b end{bmatrix}$ denote the eigenvector. Then $$ begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix} begin{bmatrix} a \ b end{bmatrix} = e^{ix} begin{bmatrix} a \ b end{bmatrix}$$ So, we get that $a = 1$ from the first equation. Solving for $b$ is my problem. I get $e^{-ix}b = e^{ix}b$, so $b = e^{2ix}b$, then $b(1 - e^{2ix}) = 0$, but then I get stuck. The eigenvalues are supposedly $$begin{bmatrix} 1 \ i end{bmatrix}, begin{bmatrix} 1 \ -i end{bmatrix}$$






linear-algebra abstract-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$




Let $x = frac{2 i pi k}{n}$ where $1 leq k leq frac{n}{2} - 1$, where $n = 2k$ even. $$ A = begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix}$$ I'm trying to find the eigenvectors of of this matrix. The eigenvalues are clearly $e^{ix}$ and $e^{-ix}$. So, let $ begin{bmatrix} a \ b end{bmatrix}$ denote the eigenvector. Then $$ begin{bmatrix} e^{ix} & 0 \ 0 & e^{-ix} end{bmatrix} begin{bmatrix} a \ b end{bmatrix} = e^{ix} begin{bmatrix} a \ b end{bmatrix}$$ So, we get that $a = 1$ from the first equation. Solving for $b$ is my problem. I get $e^{-ix}b = e^{ix}b$, so $b = e^{2ix}b$, then $b(1 - e^{2ix}) = 0$, but then I get stuck. The eigenvalues are supposedly $$begin{bmatrix} 1 \ i end{bmatrix}, begin{bmatrix} 1 \ -i end{bmatrix}$$






linear-algebra abstract-algebra eigenvalues-eigenvectors




linear-algebra abstract-algebra eigenvalues-eigenvectors






share|cite|improve this question













share|cite|improve this question











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asked Dec 1 '18 at 14:51









the manthe man

711715




711715












  • $begingroup$
    $[1, 0], [0, 1]$ always work for diagonal matrices.
    $endgroup$
    – xbh
    Dec 1 '18 at 15:01










  • $begingroup$
    $i$ appears twice and your second equation is wrong. Review what you wrote.
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:02












  • $begingroup$
    And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:07


















  • $begingroup$
    $[1, 0], [0, 1]$ always work for diagonal matrices.
    $endgroup$
    – xbh
    Dec 1 '18 at 15:01










  • $begingroup$
    $i$ appears twice and your second equation is wrong. Review what you wrote.
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:02












  • $begingroup$
    And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
    $endgroup$
    – Yves Daoust
    Dec 1 '18 at 15:07
















$begingroup$
$[1, 0], [0, 1]$ always work for diagonal matrices.
$endgroup$
– xbh
Dec 1 '18 at 15:01




$begingroup$
$[1, 0], [0, 1]$ always work for diagonal matrices.
$endgroup$
– xbh
Dec 1 '18 at 15:01












$begingroup$
$i$ appears twice and your second equation is wrong. Review what you wrote.
$endgroup$
– Yves Daoust
Dec 1 '18 at 15:02






$begingroup$
$i$ appears twice and your second equation is wrong. Review what you wrote.
$endgroup$
– Yves Daoust
Dec 1 '18 at 15:02














$begingroup$
And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
$endgroup$
– Yves Daoust
Dec 1 '18 at 15:07




$begingroup$
And I don't think the problem statement is the right one. Aren't you solving an ODE system ?
$endgroup$
– Yves Daoust
Dec 1 '18 at 15:07










2 Answers
2






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$begingroup$

Hint:



The eigenvalues are $mathrm e^{ix}$ and $mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    To find the first Eigenvector, you have to solve



    $$begin{bmatrix}e^{ix}-e^{ix}&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0\0end{bmatrix}.$$



    This is immediate.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Hint:



      The eigenvalues are $mathrm e^{ix}$ and $mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Hint:



        The eigenvalues are $mathrm e^{ix}$ and $mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint:



          The eigenvalues are $mathrm e^{ix}$ and $mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.






          share|cite|improve this answer









          $endgroup$



          Hint:



          The eigenvalues are $mathrm e^{ix}$ and $mathrm e^{-ix}$. This requires the eigenvectors have each one coordinate equal to $0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 15:01









          BernardBernard

          119k639112




          119k639112























              0












              $begingroup$

              To find the first Eigenvector, you have to solve



              $$begin{bmatrix}e^{ix}-e^{ix}&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0\0end{bmatrix}.$$



              This is immediate.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                To find the first Eigenvector, you have to solve



                $$begin{bmatrix}e^{ix}-e^{ix}&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0\0end{bmatrix}.$$



                This is immediate.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To find the first Eigenvector, you have to solve



                  $$begin{bmatrix}e^{ix}-e^{ix}&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0\0end{bmatrix}.$$



                  This is immediate.






                  share|cite|improve this answer









                  $endgroup$



                  To find the first Eigenvector, you have to solve



                  $$begin{bmatrix}e^{ix}-e^{ix}&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0&0\0&e^{-ix}-e^{ix}end{bmatrix}begin{bmatrix}a\bend{bmatrix}=begin{bmatrix}0\0end{bmatrix}.$$



                  This is immediate.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 15:05









                  Yves DaoustYves Daoust

                  125k671222




                  125k671222






























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