Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4...












1












$begingroup$


Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.



So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.



So my first question is: Can you have a submodule of rationals in a module over the integers?



If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.



(Is this correct?)






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 8:35










  • $begingroup$
    Then how do I find the basis for $L$?
    $endgroup$
    – Username Unknown
    Dec 24 '18 at 14:41






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Smith_normal_form
    $endgroup$
    – lhf
    Dec 24 '18 at 14:56
















1












$begingroup$


Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.



So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.



So my first question is: Can you have a submodule of rationals in a module over the integers?



If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.



(Is this correct?)






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 8:35










  • $begingroup$
    Then how do I find the basis for $L$?
    $endgroup$
    – Username Unknown
    Dec 24 '18 at 14:41






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Smith_normal_form
    $endgroup$
    – lhf
    Dec 24 '18 at 14:56














1












1








1





$begingroup$


Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.



So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.



So my first question is: Can you have a submodule of rationals in a module over the integers?



If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.



(Is this correct?)






linear-algebra







share|cite|improve this question











$endgroup$




Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.



So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.



So my first question is: Can you have a submodule of rationals in a module over the integers?



If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.



(Is this correct?)






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 Dec 24 '18 at 14:45







Username Unknown

















asked Dec 24 '18 at 2:02









Username UnknownUsername Unknown

1,26442158




1,26442158












  • $begingroup$
    Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 8:35










  • $begingroup$
    Then how do I find the basis for $L$?
    $endgroup$
    – Username Unknown
    Dec 24 '18 at 14:41






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Smith_normal_form
    $endgroup$
    – lhf
    Dec 24 '18 at 14:56


















  • $begingroup$
    Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 8:35










  • $begingroup$
    Then how do I find the basis for $L$?
    $endgroup$
    – Username Unknown
    Dec 24 '18 at 14:41






  • 1




    $begingroup$
    See en.wikipedia.org/wiki/Smith_normal_form
    $endgroup$
    – lhf
    Dec 24 '18 at 14:56
















$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35




$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35












$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41




$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41




1




1




$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56




$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56










1 Answer
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$begingroup$

Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.






share|cite|improve this answer











$endgroup$













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    $begingroup$

    Perform elementary column operations:
    $$
    begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
    to
    begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
    to
    begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
    $$
    Compare this with
    $$
    begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
    $$
    whose columns are a basis for $V$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Perform elementary column operations:
      $$
      begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
      to
      begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
      to
      begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
      $$
      Compare this with
      $$
      begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
      $$
      whose columns are a basis for $V$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Perform elementary column operations:
        $$
        begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
        to
        begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
        to
        begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
        $$
        Compare this with
        $$
        begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
        $$
        whose columns are a basis for $V$.






        share|cite|improve this answer











        $endgroup$



        Perform elementary column operations:
        $$
        begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
        to
        begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
        to
        begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
        $$
        Compare this with
        $$
        begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
        $$
        whose columns are a basis for $V$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 15:00

























        answered Dec 24 '18 at 14:52









        lhflhf

        165k10171396




        165k10171396






























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