Linear Algebra - Intersection of Affine Spaces












1












$begingroup$


Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.



Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.



I have found this previous question but I couldn't figure out the next steps of proving this.



We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.



I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.



Would appreciate some points and guidelines about how to approach this.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
    $endgroup$
    – SvanN
    Dec 24 '18 at 15:44






  • 1




    $begingroup$
    If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 15:53
















1












$begingroup$


Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.



Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.



I have found this previous question but I couldn't figure out the next steps of proving this.



We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.



I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.



Would appreciate some points and guidelines about how to approach this.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
    $endgroup$
    – SvanN
    Dec 24 '18 at 15:44






  • 1




    $begingroup$
    If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 15:53














1












1








1





$begingroup$


Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.



Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.



I have found this previous question but I couldn't figure out the next steps of proving this.



We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.



I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.



Would appreciate some points and guidelines about how to approach this.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$




Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.



Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.



I have found this previous question but I couldn't figure out the next steps of proving this.



We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.



I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.



Would appreciate some points and guidelines about how to approach this.






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 15:37









TegernakoTegernako

968




968








  • 1




    $begingroup$
    It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
    $endgroup$
    – SvanN
    Dec 24 '18 at 15:44






  • 1




    $begingroup$
    If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 15:53














  • 1




    $begingroup$
    It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
    $endgroup$
    – SvanN
    Dec 24 '18 at 15:44






  • 1




    $begingroup$
    If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 15:53








1




1




$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44




$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44




1




1




$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53




$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53










1 Answer
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$begingroup$

The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.






share|cite|improve this answer









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    $begingroup$

    The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
    $(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
      $(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
        $(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.






        share|cite|improve this answer









        $endgroup$



        The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
        $(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 15:53









        MirceaMircea

        1736




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