Left inverse of a matrix $3 times 2$ in $mathbb{F}_7[x]$












1












$begingroup$


Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?



I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$



begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}






matrices polynomials finite-fields inverse vector-fields







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  • $begingroup$
    Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:50


















1












$begingroup$


Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?



I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$



begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}






matrices polynomials finite-fields inverse vector-fields







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:50
















1












1








1





$begingroup$


Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?



I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$



begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}






matrices polynomials finite-fields inverse vector-fields







share|cite|improve this question











$endgroup$




Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?



I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$



begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}






matrices polynomials finite-fields inverse vector-fields




matrices polynomials finite-fields inverse vector-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 3:43









Batominovski

1




1










asked Dec 4 '18 at 2:27









Tom RyddleTom Ryddle

828




828












  • $begingroup$
    Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:50




















  • $begingroup$
    Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
    $endgroup$
    – Jyrki Lahtonen
    Dec 4 '18 at 4:50


















$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50






$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50












1 Answer
1






active

oldest

votes


















2












$begingroup$

For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.



For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$



Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$






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    $begingroup$

    For this specific case, I would make an Ansatz that a left inverse of
    $$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
    is of the form
    $$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
    where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.



    For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
    $$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
    The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
    $$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
    making
    $$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$



    Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
    The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
    $$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
    $$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$






    share|cite|improve this answer











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      2












      $begingroup$

      For this specific case, I would make an Ansatz that a left inverse of
      $$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
      is of the form
      $$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
      where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.



      For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
      $$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
      The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
      $$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
      making
      $$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$



      Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
      The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
      $$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
      $$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        For this specific case, I would make an Ansatz that a left inverse of
        $$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
        is of the form
        $$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
        where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.



        For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
        $$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
        The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
        $$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
        making
        $$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$



        Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
        The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
        $$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
        $$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$






        share|cite|improve this answer











        $endgroup$



        For this specific case, I would make an Ansatz that a left inverse of
        $$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
        is of the form
        $$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
        where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.



        For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
        $$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
        The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
        $$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
        making
        $$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$



        Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
        The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
        $$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
        $$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$







        share|cite|improve this answer














        share|cite|improve this answer



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        edited Dec 4 '18 at 13:18

























        answered Dec 4 '18 at 3:35









        BatominovskiBatominovski

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