There is a vector $y in mathbb{R^m}$ such that $A^Ty=0$ and $y^Tb not=0$












1












$begingroup$


Let there be matrix $A_{mtimes n}$ and $b in mathbb{R^m}$ , then either there exists a vector $x in mathbb{R^n}$ such that $Ax=b$ or there is a vector $y in mathbb{R^m}$ such that $A^Ty=0$ and $y^Tb not=0$



I know that if rank of of $A$ is $n$, then we will have the $1$st case :there exists a vector $x in mathbb{R^n}$ such that $Ax=b$; Now for the case rank $A <n$, how to approach? I did some: rank $A <n$ implies there is some $x not=0 in N(A)$ and we have $N(A)$ is orthogonal to $R(A^T)$. Thereafter, I need some help.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
    $endgroup$
    – Red shoes
    Jun 3 '17 at 7:52










  • $begingroup$
    Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:04










  • $begingroup$
    I know some of LPP. I want to know the proof of the $1$st comment
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:37










  • $begingroup$
    Never mind, I'll write an elementary proof below
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:49
















1












$begingroup$


Let there be matrix $A_{mtimes n}$ and $b in mathbb{R^m}$ , then either there exists a vector $x in mathbb{R^n}$ such that $Ax=b$ or there is a vector $y in mathbb{R^m}$ such that $A^Ty=0$ and $y^Tb not=0$



I know that if rank of of $A$ is $n$, then we will have the $1$st case :there exists a vector $x in mathbb{R^n}$ such that $Ax=b$; Now for the case rank $A <n$, how to approach? I did some: rank $A <n$ implies there is some $x not=0 in N(A)$ and we have $N(A)$ is orthogonal to $R(A^T)$. Thereafter, I need some help.






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
    $endgroup$
    – Red shoes
    Jun 3 '17 at 7:52










  • $begingroup$
    Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:04










  • $begingroup$
    I know some of LPP. I want to know the proof of the $1$st comment
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:37










  • $begingroup$
    Never mind, I'll write an elementary proof below
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:49














1












1








1


1



$begingroup$


Let there be matrix $A_{mtimes n}$ and $b in mathbb{R^m}$ , then either there exists a vector $x in mathbb{R^n}$ such that $Ax=b$ or there is a vector $y in mathbb{R^m}$ such that $A^Ty=0$ and $y^Tb not=0$



I know that if rank of of $A$ is $n$, then we will have the $1$st case :there exists a vector $x in mathbb{R^n}$ such that $Ax=b$; Now for the case rank $A <n$, how to approach? I did some: rank $A <n$ implies there is some $x not=0 in N(A)$ and we have $N(A)$ is orthogonal to $R(A^T)$. Thereafter, I need some help.






linear-algebra matrices







share|cite|improve this question











$endgroup$




Let there be matrix $A_{mtimes n}$ and $b in mathbb{R^m}$ , then either there exists a vector $x in mathbb{R^n}$ such that $Ax=b$ or there is a vector $y in mathbb{R^m}$ such that $A^Ty=0$ and $y^Tb not=0$



I know that if rank of of $A$ is $n$, then we will have the $1$st case :there exists a vector $x in mathbb{R^n}$ such that $Ax=b$; Now for the case rank $A <n$, how to approach? I did some: rank $A <n$ implies there is some $x not=0 in N(A)$ and we have $N(A)$ is orthogonal to $R(A^T)$. Thereafter, I need some help.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 10:50









José Carlos Santos

157k22126227




157k22126227










asked Jun 3 '17 at 7:22









Hirakjyoti DasHirakjyoti Das

389112




389112












  • $begingroup$
    if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
    $endgroup$
    – Red shoes
    Jun 3 '17 at 7:52










  • $begingroup$
    Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:04










  • $begingroup$
    I know some of LPP. I want to know the proof of the $1$st comment
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:37










  • $begingroup$
    Never mind, I'll write an elementary proof below
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:49


















  • $begingroup$
    if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
    $endgroup$
    – Red shoes
    Jun 3 '17 at 7:52










  • $begingroup$
    Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:04










  • $begingroup$
    I know some of LPP. I want to know the proof of the $1$st comment
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:37










  • $begingroup$
    Never mind, I'll write an elementary proof below
    $endgroup$
    – Red shoes
    Jun 3 '17 at 8:49
















$begingroup$
if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
$endgroup$
– Red shoes
Jun 3 '17 at 7:52




$begingroup$
if rank of $A$ is $n$, still you can't guarantee $Ax=b$ has solution!! if rank $A$ =m then $Ax=b$ always has solution
$endgroup$
– Red shoes
Jun 3 '17 at 7:52












$begingroup$
Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
$endgroup$
– Red shoes
Jun 3 '17 at 8:04




$begingroup$
Are you familiar with linear programing and duality or Alternative theorems? If yes , I will solve this for you in less than three lines
$endgroup$
– Red shoes
Jun 3 '17 at 8:04












$begingroup$
I know some of LPP. I want to know the proof of the $1$st comment
$endgroup$
– Hirakjyoti Das
Jun 3 '17 at 8:37




$begingroup$
I know some of LPP. I want to know the proof of the $1$st comment
$endgroup$
– Hirakjyoti Das
Jun 3 '17 at 8:37












$begingroup$
Never mind, I'll write an elementary proof below
$endgroup$
– Red shoes
Jun 3 '17 at 8:49




$begingroup$
Never mind, I'll write an elementary proof below
$endgroup$
– Red shoes
Jun 3 '17 at 8:49










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $ Ax=b$ has no solution, thus $b notin R(A).$ we know that $R^m = R(A)^{perp} + R(A)$. So $$b=y+Ax$$ for some nonzero $y in R(A)^{perp}=N(A^T) $ and some $xin R^n$. Hence ; $$A^Ty=0$$ and $$y^Tb=|y|^2+y^TAx = |y|^2 neq 0$$
.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that's very helpful, thank you
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:55





















0












$begingroup$

Imagine that you try to solve the equation $Ax=b$. You can do it by Gaussian elimination. In the end, you get an equation of the type $A'x=b'$, whre $A'$ is something like$$begin{pmatrix}alpha_1&*&*&*&*&*\0&0&alpha_2&*&*&*\0&0&0&alpha_3&*&*\0&0&0&0&0&0\0&0&0&0&0&0end{pmatrix},$$with every $alpha_kneq0$. So, the system has a solution if and only if the last two coordinates of $b'$ are equal to $0$. Otherwise, take $y=(0,0,0,beta_4,beta_5)$, where $beta_4$ and $beta_5$ are the two last coordinates of $b'$.






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









    2












    $begingroup$

    Let $ Ax=b$ has no solution, thus $b notin R(A).$ we know that $R^m = R(A)^{perp} + R(A)$. So $$b=y+Ax$$ for some nonzero $y in R(A)^{perp}=N(A^T) $ and some $xin R^n$. Hence ; $$A^Ty=0$$ and $$y^Tb=|y|^2+y^TAx = |y|^2 neq 0$$
    .






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      that's very helpful, thank you
      $endgroup$
      – Hirakjyoti Das
      Jun 3 '17 at 8:55


















    2












    $begingroup$

    Let $ Ax=b$ has no solution, thus $b notin R(A).$ we know that $R^m = R(A)^{perp} + R(A)$. So $$b=y+Ax$$ for some nonzero $y in R(A)^{perp}=N(A^T) $ and some $xin R^n$. Hence ; $$A^Ty=0$$ and $$y^Tb=|y|^2+y^TAx = |y|^2 neq 0$$
    .






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      that's very helpful, thank you
      $endgroup$
      – Hirakjyoti Das
      Jun 3 '17 at 8:55
















    2












    2








    2





    $begingroup$

    Let $ Ax=b$ has no solution, thus $b notin R(A).$ we know that $R^m = R(A)^{perp} + R(A)$. So $$b=y+Ax$$ for some nonzero $y in R(A)^{perp}=N(A^T) $ and some $xin R^n$. Hence ; $$A^Ty=0$$ and $$y^Tb=|y|^2+y^TAx = |y|^2 neq 0$$
    .






    share|cite|improve this answer











    $endgroup$



    Let $ Ax=b$ has no solution, thus $b notin R(A).$ we know that $R^m = R(A)^{perp} + R(A)$. So $$b=y+Ax$$ for some nonzero $y in R(A)^{perp}=N(A^T) $ and some $xin R^n$. Hence ; $$A^Ty=0$$ and $$y^Tb=|y|^2+y^TAx = |y|^2 neq 0$$
    .







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 3 '17 at 8:53

























    answered Jun 3 '17 at 8:45









    Red shoesRed shoes

    4,744621




    4,744621












    • $begingroup$
      that's very helpful, thank you
      $endgroup$
      – Hirakjyoti Das
      Jun 3 '17 at 8:55




















    • $begingroup$
      that's very helpful, thank you
      $endgroup$
      – Hirakjyoti Das
      Jun 3 '17 at 8:55


















    $begingroup$
    that's very helpful, thank you
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:55






    $begingroup$
    that's very helpful, thank you
    $endgroup$
    – Hirakjyoti Das
    Jun 3 '17 at 8:55













    0












    $begingroup$

    Imagine that you try to solve the equation $Ax=b$. You can do it by Gaussian elimination. In the end, you get an equation of the type $A'x=b'$, whre $A'$ is something like$$begin{pmatrix}alpha_1&*&*&*&*&*\0&0&alpha_2&*&*&*\0&0&0&alpha_3&*&*\0&0&0&0&0&0\0&0&0&0&0&0end{pmatrix},$$with every $alpha_kneq0$. So, the system has a solution if and only if the last two coordinates of $b'$ are equal to $0$. Otherwise, take $y=(0,0,0,beta_4,beta_5)$, where $beta_4$ and $beta_5$ are the two last coordinates of $b'$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Imagine that you try to solve the equation $Ax=b$. You can do it by Gaussian elimination. In the end, you get an equation of the type $A'x=b'$, whre $A'$ is something like$$begin{pmatrix}alpha_1&*&*&*&*&*\0&0&alpha_2&*&*&*\0&0&0&alpha_3&*&*\0&0&0&0&0&0\0&0&0&0&0&0end{pmatrix},$$with every $alpha_kneq0$. So, the system has a solution if and only if the last two coordinates of $b'$ are equal to $0$. Otherwise, take $y=(0,0,0,beta_4,beta_5)$, where $beta_4$ and $beta_5$ are the two last coordinates of $b'$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Imagine that you try to solve the equation $Ax=b$. You can do it by Gaussian elimination. In the end, you get an equation of the type $A'x=b'$, whre $A'$ is something like$$begin{pmatrix}alpha_1&*&*&*&*&*\0&0&alpha_2&*&*&*\0&0&0&alpha_3&*&*\0&0&0&0&0&0\0&0&0&0&0&0end{pmatrix},$$with every $alpha_kneq0$. So, the system has a solution if and only if the last two coordinates of $b'$ are equal to $0$. Otherwise, take $y=(0,0,0,beta_4,beta_5)$, where $beta_4$ and $beta_5$ are the two last coordinates of $b'$.






        share|cite|improve this answer









        $endgroup$



        Imagine that you try to solve the equation $Ax=b$. You can do it by Gaussian elimination. In the end, you get an equation of the type $A'x=b'$, whre $A'$ is something like$$begin{pmatrix}alpha_1&*&*&*&*&*\0&0&alpha_2&*&*&*\0&0&0&alpha_3&*&*\0&0&0&0&0&0\0&0&0&0&0&0end{pmatrix},$$with every $alpha_kneq0$. So, the system has a solution if and only if the last two coordinates of $b'$ are equal to $0$. Otherwise, take $y=(0,0,0,beta_4,beta_5)$, where $beta_4$ and $beta_5$ are the two last coordinates of $b'$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 3 '17 at 8:07









        José Carlos SantosJosé Carlos Santos

        157k22126227




        157k22126227






























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