Kernel of product [closed]












1












$begingroup$


Let $B in mathbb R^{{k}times {m}}$ and $A in mathbb R^{{m}times {n}}$. Further assume that $operatorname{ker}(B) cap operatorname{ran}(A) = {0}.$ Show that this implies $operatorname{ker}(A) = operatorname{ker}(BA).$



I have no idea how to tackle this problem.






matrices







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closed as off-topic by Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister Nov 30 '18 at 15:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    math.meta.stackexchange.com/questions/9959/…
    $endgroup$
    – Trevor Gunn
    Nov 30 '18 at 13:04










  • $begingroup$
    What is Ran(A)?
    $endgroup$
    – Bernard
    Nov 30 '18 at 13:14










  • $begingroup$
    Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
    $endgroup$
    – Eric
    Nov 30 '18 at 13:16








  • 1




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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    – José Carlos Santos
    Nov 30 '18 at 13:20
















1












$begingroup$


Let $B in mathbb R^{{k}times {m}}$ and $A in mathbb R^{{m}times {n}}$. Further assume that $operatorname{ker}(B) cap operatorname{ran}(A) = {0}.$ Show that this implies $operatorname{ker}(A) = operatorname{ker}(BA).$



I have no idea how to tackle this problem.






matrices







share|cite|improve this question











$endgroup$



closed as off-topic by Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister Nov 30 '18 at 15:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    math.meta.stackexchange.com/questions/9959/…
    $endgroup$
    – Trevor Gunn
    Nov 30 '18 at 13:04










  • $begingroup$
    What is Ran(A)?
    $endgroup$
    – Bernard
    Nov 30 '18 at 13:14










  • $begingroup$
    Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
    $endgroup$
    – Eric
    Nov 30 '18 at 13:16








  • 1




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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    – José Carlos Santos
    Nov 30 '18 at 13:20














1












1








1





$begingroup$


Let $B in mathbb R^{{k}times {m}}$ and $A in mathbb R^{{m}times {n}}$. Further assume that $operatorname{ker}(B) cap operatorname{ran}(A) = {0}.$ Show that this implies $operatorname{ker}(A) = operatorname{ker}(BA).$



I have no idea how to tackle this problem.






matrices







share|cite|improve this question











$endgroup$




Let $B in mathbb R^{{k}times {m}}$ and $A in mathbb R^{{m}times {n}}$. Further assume that $operatorname{ker}(B) cap operatorname{ran}(A) = {0}.$ Show that this implies $operatorname{ker}(A) = operatorname{ker}(BA).$



I have no idea how to tackle this problem.






matrices




matrices






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 14:17









amWhy

1




1










asked Nov 30 '18 at 13:02









Luke3Luke3

63




63




closed as off-topic by Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister Nov 30 '18 at 15:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister Nov 30 '18 at 15:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Trevor Gunn, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, RRL, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    math.meta.stackexchange.com/questions/9959/…
    $endgroup$
    – Trevor Gunn
    Nov 30 '18 at 13:04










  • $begingroup$
    What is Ran(A)?
    $endgroup$
    – Bernard
    Nov 30 '18 at 13:14










  • $begingroup$
    Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
    $endgroup$
    – Eric
    Nov 30 '18 at 13:16








  • 1




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Nov 30 '18 at 13:20














  • 1




    $begingroup$
    math.meta.stackexchange.com/questions/9959/…
    $endgroup$
    – Trevor Gunn
    Nov 30 '18 at 13:04










  • $begingroup$
    What is Ran(A)?
    $endgroup$
    – Bernard
    Nov 30 '18 at 13:14










  • $begingroup$
    Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
    $endgroup$
    – Eric
    Nov 30 '18 at 13:16








  • 1




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Nov 30 '18 at 13:20








1




1




$begingroup$
math.meta.stackexchange.com/questions/9959/…
$endgroup$
– Trevor Gunn
Nov 30 '18 at 13:04




$begingroup$
math.meta.stackexchange.com/questions/9959/…
$endgroup$
– Trevor Gunn
Nov 30 '18 at 13:04












$begingroup$
What is Ran(A)?
$endgroup$
– Bernard
Nov 30 '18 at 13:14




$begingroup$
What is Ran(A)?
$endgroup$
– Bernard
Nov 30 '18 at 13:14












$begingroup$
Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
$endgroup$
– Eric
Nov 30 '18 at 13:16






$begingroup$
Suppose $BAx=0$. What can you say about the vector $Ax$ in terms of $mathrm{Ran}(A)$ and $mathrm{Ker}(B)$?
$endgroup$
– Eric
Nov 30 '18 at 13:16






1




1




$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 13:20




$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 30 '18 at 13:20










2 Answers
2






active

oldest

votes


















1












$begingroup$

$ker(A)subseteqker(BA)$ is obvious.



Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is not hard, but the structure is a bit verbose, hope that helps.
    $$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
    $$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
    If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$



    or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
    $$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
    But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$



    If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
    $$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
      $endgroup$
      – Eric
      Nov 30 '18 at 14:24












    • $begingroup$
      You are right @Eric, I'll edit that.
      $endgroup$
      – Eduardo Elael
      Nov 30 '18 at 15:30




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $ker(A)subseteqker(BA)$ is obvious.



    Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $ker(A)subseteqker(BA)$ is obvious.



      Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $ker(A)subseteqker(BA)$ is obvious.



        Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.






        share|cite|improve this answer









        $endgroup$



        $ker(A)subseteqker(BA)$ is obvious.



        Suppose $vinker(BA)$. Then $B(Av)=0$, so $Avinker(B)capoperatorname{ran}(A)$. Hence $Av=0$, so $vinker(A)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 15:12









        egregegreg

        180k1485202




        180k1485202























            0












            $begingroup$

            It is not hard, but the structure is a bit verbose, hope that helps.
            $$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
            $$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
            If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$



            or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
            $$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
            But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$



            If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
            $$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
              $endgroup$
              – Eric
              Nov 30 '18 at 14:24












            • $begingroup$
              You are right @Eric, I'll edit that.
              $endgroup$
              – Eduardo Elael
              Nov 30 '18 at 15:30


















            0












            $begingroup$

            It is not hard, but the structure is a bit verbose, hope that helps.
            $$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
            $$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
            If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$



            or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
            $$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
            But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$



            If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
            $$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
              $endgroup$
              – Eric
              Nov 30 '18 at 14:24












            • $begingroup$
              You are right @Eric, I'll edit that.
              $endgroup$
              – Eduardo Elael
              Nov 30 '18 at 15:30
















            0












            0








            0





            $begingroup$

            It is not hard, but the structure is a bit verbose, hope that helps.
            $$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
            $$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
            If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$



            or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
            $$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
            But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$



            If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
            $$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$






            share|cite|improve this answer











            $endgroup$



            It is not hard, but the structure is a bit verbose, hope that helps.
            $$text{Ker}(A)={v in mathbb{R}^n ~|~Av=0}$$
            $$text{Ker}(BA)={v in mathbb{R}^n ~|~BAv=0}$$
            If $vintext{Ker}(A)Rightarrow Av=0 Rightarrow BAv=0Rightarrow vintext{Ker}(BA)$



            or if $Av=uneq0Rightarrow (uintext{Ran}(A)$ and $B(Av)=0 iff Bu=0)$, thus:
            $$text{Ker}(BA)= text{Ker}(A) cup {vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }$$
            But, $text{Ker}(B)captext{Ran}(A)={0} Rightarrow (uintext{Ran}(A)/{0}Rightarrow unotintext{Ker}(B))$



            If $vin mathbb{R}^n/text{Ker}(A) Rightarrow Avintext{Ran}(A)/{0}$, thus (from above) $unotin text{Ker}(B)$, so no element of $vin mathbb{R}^n/text{Ker}(A)$ satisfies $uin text{Ker}(B)$ and ${vin mathbb{R}^n/text{Ker}(A)~|~u=Av~text{and}~uintext{Ker}(B) }={}$, finally:
            $$text{Ker}(BA)= text{Ker}(A) cup {} = text{Ker}(A)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 30 '18 at 15:30

























            answered Nov 30 '18 at 13:57









            Eduardo ElaelEduardo Elael

            30115




            30115












            • $begingroup$
              $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
              $endgroup$
              – Eric
              Nov 30 '18 at 14:24












            • $begingroup$
              You are right @Eric, I'll edit that.
              $endgroup$
              – Eduardo Elael
              Nov 30 '18 at 15:30




















            • $begingroup$
              $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
              $endgroup$
              – Eric
              Nov 30 '18 at 14:24












            • $begingroup$
              You are right @Eric, I'll edit that.
              $endgroup$
              – Eduardo Elael
              Nov 30 '18 at 15:30


















            $begingroup$
            $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
            $endgroup$
            – Eric
            Nov 30 '18 at 14:24






            $begingroup$
            $0$ is always an element of the kernel of a matrix. You should consider $v in mathbb{R}^n $ in your definition, not $v in mathbb{R}^n backslash {0}$.
            $endgroup$
            – Eric
            Nov 30 '18 at 14:24














            $begingroup$
            You are right @Eric, I'll edit that.
            $endgroup$
            – Eduardo Elael
            Nov 30 '18 at 15:30






            $begingroup$
            You are right @Eric, I'll edit that.
            $endgroup$
            – Eduardo Elael
            Nov 30 '18 at 15:30





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