finding the eigenvectors of a matrix












0












$begingroup$


let be a matrix $textbf{A}$



$$
textbf{A} =
begin{bmatrix}
0 & 0 & 0 \
0 & 0 & frac{xalpha}{2H} \
0 & frac{xalpha}{2H} & 0 \
end{bmatrix}
$$



The eigenvalues of $textbf{A}$ are given by



$$
det(textbf{A} - lambdamathbb{I}) = det begin{bmatrix}
-lambda & 0 & 0 \
0 & -lambda & frac{xalpha}{2H} \
0 & frac{xalpha}{2H} & -lambda \
end{bmatrix} = -lambda(lambda^2-(frac{xalpha}{2H})^2)=0
$$
hence we find:



$$lambda_{1}=0 quad lambda_{2,3} = pm frac{xalpha}{2H}
$$



In order to find the eigenvectors, we have to find the solution to



$$
begin{bmatrix}
-lambda & 0 & 0 \
0 & -lambda & frac{xalpha}{2H} \
0 & frac{xalpha}{2H} & -lambda \
end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix}
$$



for each of the three values of $lambda$.



Thus we find the second eigenvalues $$vec{e}_2 = begin{pmatrix}0 \ 1 \1end{pmatrix}
$$
and the third eigenvalue
$$vec{e}_3 = begin{pmatrix}0 \ 1 \-1end{pmatrix}
$$



Here is my question:



In the solution of the exercice I have, I'm given the first eigenvector as



$$vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}
$$



However, I would tend to say that the solution $vec{e}_1$ with eigenvalue $lambda_1 = 0$ is given by



$$vec{e}_1 = begin{pmatrix}0 \ 0 \0end{pmatrix}
$$



What am I missing?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    let be a matrix $textbf{A}$



    $$
    textbf{A} =
    begin{bmatrix}
    0 & 0 & 0 \
    0 & 0 & frac{xalpha}{2H} \
    0 & frac{xalpha}{2H} & 0 \
    end{bmatrix}
    $$



    The eigenvalues of $textbf{A}$ are given by



    $$
    det(textbf{A} - lambdamathbb{I}) = det begin{bmatrix}
    -lambda & 0 & 0 \
    0 & -lambda & frac{xalpha}{2H} \
    0 & frac{xalpha}{2H} & -lambda \
    end{bmatrix} = -lambda(lambda^2-(frac{xalpha}{2H})^2)=0
    $$
    hence we find:



    $$lambda_{1}=0 quad lambda_{2,3} = pm frac{xalpha}{2H}
    $$



    In order to find the eigenvectors, we have to find the solution to



    $$
    begin{bmatrix}
    -lambda & 0 & 0 \
    0 & -lambda & frac{xalpha}{2H} \
    0 & frac{xalpha}{2H} & -lambda \
    end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix}
    $$



    for each of the three values of $lambda$.



    Thus we find the second eigenvalues $$vec{e}_2 = begin{pmatrix}0 \ 1 \1end{pmatrix}
    $$
    and the third eigenvalue
    $$vec{e}_3 = begin{pmatrix}0 \ 1 \-1end{pmatrix}
    $$



    Here is my question:



    In the solution of the exercice I have, I'm given the first eigenvector as



    $$vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}
    $$



    However, I would tend to say that the solution $vec{e}_1$ with eigenvalue $lambda_1 = 0$ is given by



    $$vec{e}_1 = begin{pmatrix}0 \ 0 \0end{pmatrix}
    $$



    What am I missing?






    linear-algebra matrices eigenvalues-eigenvectors







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      let be a matrix $textbf{A}$



      $$
      textbf{A} =
      begin{bmatrix}
      0 & 0 & 0 \
      0 & 0 & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & 0 \
      end{bmatrix}
      $$



      The eigenvalues of $textbf{A}$ are given by



      $$
      det(textbf{A} - lambdamathbb{I}) = det begin{bmatrix}
      -lambda & 0 & 0 \
      0 & -lambda & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & -lambda \
      end{bmatrix} = -lambda(lambda^2-(frac{xalpha}{2H})^2)=0
      $$
      hence we find:



      $$lambda_{1}=0 quad lambda_{2,3} = pm frac{xalpha}{2H}
      $$



      In order to find the eigenvectors, we have to find the solution to



      $$
      begin{bmatrix}
      -lambda & 0 & 0 \
      0 & -lambda & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & -lambda \
      end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix}
      $$



      for each of the three values of $lambda$.



      Thus we find the second eigenvalues $$vec{e}_2 = begin{pmatrix}0 \ 1 \1end{pmatrix}
      $$
      and the third eigenvalue
      $$vec{e}_3 = begin{pmatrix}0 \ 1 \-1end{pmatrix}
      $$



      Here is my question:



      In the solution of the exercice I have, I'm given the first eigenvector as



      $$vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}
      $$



      However, I would tend to say that the solution $vec{e}_1$ with eigenvalue $lambda_1 = 0$ is given by



      $$vec{e}_1 = begin{pmatrix}0 \ 0 \0end{pmatrix}
      $$



      What am I missing?






      linear-algebra matrices eigenvalues-eigenvectors







      share|cite|improve this question











      $endgroup$




      let be a matrix $textbf{A}$



      $$
      textbf{A} =
      begin{bmatrix}
      0 & 0 & 0 \
      0 & 0 & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & 0 \
      end{bmatrix}
      $$



      The eigenvalues of $textbf{A}$ are given by



      $$
      det(textbf{A} - lambdamathbb{I}) = det begin{bmatrix}
      -lambda & 0 & 0 \
      0 & -lambda & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & -lambda \
      end{bmatrix} = -lambda(lambda^2-(frac{xalpha}{2H})^2)=0
      $$
      hence we find:



      $$lambda_{1}=0 quad lambda_{2,3} = pm frac{xalpha}{2H}
      $$



      In order to find the eigenvectors, we have to find the solution to



      $$
      begin{bmatrix}
      -lambda & 0 & 0 \
      0 & -lambda & frac{xalpha}{2H} \
      0 & frac{xalpha}{2H} & -lambda \
      end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix}
      $$



      for each of the three values of $lambda$.



      Thus we find the second eigenvalues $$vec{e}_2 = begin{pmatrix}0 \ 1 \1end{pmatrix}
      $$
      and the third eigenvalue
      $$vec{e}_3 = begin{pmatrix}0 \ 1 \-1end{pmatrix}
      $$



      Here is my question:



      In the solution of the exercice I have, I'm given the first eigenvector as



      $$vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}
      $$



      However, I would tend to say that the solution $vec{e}_1$ with eigenvalue $lambda_1 = 0$ is given by



      $$vec{e}_1 = begin{pmatrix}0 \ 0 \0end{pmatrix}
      $$



      What am I missing?






      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 Dec 23 '18 at 22:38









      user376343

      3,9584829




      3,9584829










      asked Dec 23 '18 at 22:31









      ecjbecjb

      2858




      2858






















          1 Answer
          1






          active

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          5












          $begingroup$

          You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $lambda$ is a non-null vector $v$ such that $A.v=lambda v$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
            $endgroup$
            – ecjb
            Dec 23 '18 at 22:50












          • $begingroup$
            Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 22:58












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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $lambda$ is a non-null vector $v$ such that $A.v=lambda v$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
            $endgroup$
            – ecjb
            Dec 23 '18 at 22:50












          • $begingroup$
            Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 22:58
















          5












          $begingroup$

          You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $lambda$ is a non-null vector $v$ such that $A.v=lambda v$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
            $endgroup$
            – ecjb
            Dec 23 '18 at 22:50












          • $begingroup$
            Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 22:58














          5












          5








          5





          $begingroup$

          You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $lambda$ is a non-null vector $v$ such that $A.v=lambda v$.






          share|cite|improve this answer









          $endgroup$



          You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $lambda$ is a non-null vector $v$ such that $A.v=lambda v$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 22:34









          José Carlos SantosJosé Carlos Santos

          172k23132240




          172k23132240












          • $begingroup$
            Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
            $endgroup$
            – ecjb
            Dec 23 '18 at 22:50












          • $begingroup$
            Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 22:58


















          • $begingroup$
            Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
            $endgroup$
            – ecjb
            Dec 23 '18 at 22:50












          • $begingroup$
            Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
            $endgroup$
            – José Carlos Santos
            Dec 23 '18 at 22:58
















          $begingroup$
          Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
          $endgroup$
          – ecjb
          Dec 23 '18 at 22:50






          $begingroup$
          Thank you @Jose Carlos Santos. So $begin{pmatrix}0 \ 0 \0end{pmatrix} $ cannot be an eigenvector. Now, if we compute $lambda_1 = 0$ in $ begin{bmatrix} -lambda & 0 & 0 \ 0 & -lambda & frac{xalpha}{2H} \ 0 & frac{xalpha}{2H} & -lambda \ end{bmatrix}begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 0 \ 0 \ 0 end{pmatrix} $, how can we obtain the first eigenvector $vec{e}_1 = begin{pmatrix}1 \ 0 \0end{pmatrix}$?
          $endgroup$
          – ecjb
          Dec 23 '18 at 22:50














          $begingroup$
          Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 22:58




          $begingroup$
          Solve the system $A.begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$.
          $endgroup$
          – José Carlos Santos
          Dec 23 '18 at 22:58


















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