Writing in vector form the set of solutions giving a matrix












-2












$begingroup$


Write in vector form the set of solutions to the homogeneous system of linear equations with matrix $A$, defined by:



begin{bmatrix}
0& 0\
0& 0\
end{bmatrix}



This is a $2 times 2$ matrix which has all entries are zero.



$A$ is the homogeneous system that is given to us. It is also a $2 times 2$ matrix which has all entries equal to zero). But, I did not understand this question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:14










  • $begingroup$
    yes. I guess it ..
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:16










  • $begingroup$
    Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:18










  • $begingroup$
    All vectors ? The answer is R ?
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:19










  • $begingroup$
    Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:29
















-2












$begingroup$


Write in vector form the set of solutions to the homogeneous system of linear equations with matrix $A$, defined by:



begin{bmatrix}
0& 0\
0& 0\
end{bmatrix}



This is a $2 times 2$ matrix which has all entries are zero.



$A$ is the homogeneous system that is given to us. It is also a $2 times 2$ matrix which has all entries equal to zero). But, I did not understand this question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:14










  • $begingroup$
    yes. I guess it ..
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:16










  • $begingroup$
    Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:18










  • $begingroup$
    All vectors ? The answer is R ?
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:19










  • $begingroup$
    Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:29














-2












-2








-2





$begingroup$


Write in vector form the set of solutions to the homogeneous system of linear equations with matrix $A$, defined by:



begin{bmatrix}
0& 0\
0& 0\
end{bmatrix}



This is a $2 times 2$ matrix which has all entries are zero.



$A$ is the homogeneous system that is given to us. It is also a $2 times 2$ matrix which has all entries equal to zero). But, I did not understand this question.






linear-algebra







share|cite|improve this question











$endgroup$




Write in vector form the set of solutions to the homogeneous system of linear equations with matrix $A$, defined by:



begin{bmatrix}
0& 0\
0& 0\
end{bmatrix}



This is a $2 times 2$ matrix which has all entries are zero.



$A$ is the homogeneous system that is given to us. It is also a $2 times 2$ matrix which has all entries equal to zero). But, I did not understand this question.






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 31 '18 at 0:33









Gaby Alfonso

1,1811318




1,1811318










asked Dec 31 '18 at 0:11









EfemisaktanEfemisaktan

33




33












  • $begingroup$
    So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:14










  • $begingroup$
    yes. I guess it ..
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:16










  • $begingroup$
    Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:18










  • $begingroup$
    All vectors ? The answer is R ?
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:19










  • $begingroup$
    Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:29


















  • $begingroup$
    So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:14










  • $begingroup$
    yes. I guess it ..
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:16










  • $begingroup$
    Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
    $endgroup$
    – Dave
    Dec 31 '18 at 0:18










  • $begingroup$
    All vectors ? The answer is R ?
    $endgroup$
    – Efemisaktan
    Dec 31 '18 at 0:19










  • $begingroup$
    Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:29
















$begingroup$
So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
$endgroup$
– Dave
Dec 31 '18 at 0:14




$begingroup$
So $A$ is the $2times 2$ zero matrix and you want the solutions to the homogeneous system $Ax=0$?
$endgroup$
– Dave
Dec 31 '18 at 0:14












$begingroup$
yes. I guess it ..
$endgroup$
– Efemisaktan
Dec 31 '18 at 0:16




$begingroup$
yes. I guess it ..
$endgroup$
– Efemisaktan
Dec 31 '18 at 0:16












$begingroup$
Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
$endgroup$
– Dave
Dec 31 '18 at 0:18




$begingroup$
Well what vectors $x$ can you multiply by the zero matrix to get the zero vector?
$endgroup$
– Dave
Dec 31 '18 at 0:18












$begingroup$
All vectors ? The answer is R ?
$endgroup$
– Efemisaktan
Dec 31 '18 at 0:19




$begingroup$
All vectors ? The answer is R ?
$endgroup$
– Efemisaktan
Dec 31 '18 at 0:19












$begingroup$
Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
$endgroup$
– Dave
Dec 31 '18 at 0:29




$begingroup$
Well yeah it's all vectors, but that's not $mathbb R$. If your ground field is $mathbb R$, then it would be all vectors in $mathbb R^2$ that solve $Ax=0$.
$endgroup$
– Dave
Dec 31 '18 at 0:29










2 Answers
2






active

oldest

votes


















0












$begingroup$

If you have the system $$left(begin{matrix} 0 & 0 \ 0 & 0 end{matrix}right)x=left(begin{matrix}0 \ 0 end{matrix}right)$$ the solution is $mathbb{R}^2$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    why the down vote?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 0:21



















0












$begingroup$

In case you don't want the intuitive answer (as in the comments).



Clearly, the domain of this linear mapping is in the $R^2$ space. Since the rank of this matrix is 0, the null space is also dimension 2 which spans the entire same $R^2$ vector space.






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









    0












    $begingroup$

    If you have the system $$left(begin{matrix} 0 & 0 \ 0 & 0 end{matrix}right)x=left(begin{matrix}0 \ 0 end{matrix}right)$$ the solution is $mathbb{R}^2$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      why the down vote?
      $endgroup$
      – Martín Vacas Vignolo
      Dec 31 '18 at 0:21
















    0












    $begingroup$

    If you have the system $$left(begin{matrix} 0 & 0 \ 0 & 0 end{matrix}right)x=left(begin{matrix}0 \ 0 end{matrix}right)$$ the solution is $mathbb{R}^2$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      why the down vote?
      $endgroup$
      – Martín Vacas Vignolo
      Dec 31 '18 at 0:21














    0












    0








    0





    $begingroup$

    If you have the system $$left(begin{matrix} 0 & 0 \ 0 & 0 end{matrix}right)x=left(begin{matrix}0 \ 0 end{matrix}right)$$ the solution is $mathbb{R}^2$






    share|cite|improve this answer









    $endgroup$



    If you have the system $$left(begin{matrix} 0 & 0 \ 0 & 0 end{matrix}right)x=left(begin{matrix}0 \ 0 end{matrix}right)$$ the solution is $mathbb{R}^2$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 31 '18 at 0:15









    Martín Vacas VignoloMartín Vacas Vignolo

    3,816623




    3,816623








    • 2




      $begingroup$
      why the down vote?
      $endgroup$
      – Martín Vacas Vignolo
      Dec 31 '18 at 0:21














    • 2




      $begingroup$
      why the down vote?
      $endgroup$
      – Martín Vacas Vignolo
      Dec 31 '18 at 0:21








    2




    2




    $begingroup$
    why the down vote?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 0:21




    $begingroup$
    why the down vote?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 0:21











    0












    $begingroup$

    In case you don't want the intuitive answer (as in the comments).



    Clearly, the domain of this linear mapping is in the $R^2$ space. Since the rank of this matrix is 0, the null space is also dimension 2 which spans the entire same $R^2$ vector space.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In case you don't want the intuitive answer (as in the comments).



      Clearly, the domain of this linear mapping is in the $R^2$ space. Since the rank of this matrix is 0, the null space is also dimension 2 which spans the entire same $R^2$ vector space.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In case you don't want the intuitive answer (as in the comments).



        Clearly, the domain of this linear mapping is in the $R^2$ space. Since the rank of this matrix is 0, the null space is also dimension 2 which spans the entire same $R^2$ vector space.






        share|cite|improve this answer









        $endgroup$



        In case you don't want the intuitive answer (as in the comments).



        Clearly, the domain of this linear mapping is in the $R^2$ space. Since the rank of this matrix is 0, the null space is also dimension 2 which spans the entire same $R^2$ vector space.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 0:43









        Wangkun XuWangkun Xu

        485




        485






























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