Show that intersection of a polyhedron and affine set is a polyhedron.












1












$begingroup$


Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.



To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is



$$
L= {v: cv=0}
$$



$$
u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
$$

where $d=Cu$.



To show the claim I need to assume $x$ is in $P$, so



$$
Ax geq b
$$



Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.



Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
How can I proceed?










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$endgroup$

















    1












    $begingroup$


    Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
    Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.



    To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is



    $$
    L= {v: cv=0}
    $$



    $$
    u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
    $$

    where $d=Cu$.



    To show the claim I need to assume $x$ is in $P$, so



    $$
    Ax geq b
    $$



    Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.



    Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
    How can I proceed?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
      Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.



      To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is



      $$
      L= {v: cv=0}
      $$



      $$
      u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
      $$

      where $d=Cu$.



      To show the claim I need to assume $x$ is in $P$, so



      $$
      Ax geq b
      $$



      Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.



      Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
      How can I proceed?










      share|cite|improve this question











      $endgroup$




      Let $P={x in mathbb{R}^n mid Ax geq b}$ be a nonempty polyhedron for a matrix $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.
      Show that a nonempty intersection of $P$ and an affine set in $mathbb{R}^{n}$ is a polyhedron.



      To show this I just know that an affine set in $mathbb{R}^{n}$ can be written as a subspace $L$ in $mathbb{R}^{n}$ plus some $u$ in $mathbb{R}^{n}$, that is



      $$
      L= {v: cv=0}
      $$



      $$
      u+L={u+v mid v in N(C)} ,,,text{or},,, u+L = {z mid Cz=d}
      $$

      where $d=Cu$.



      To show the claim I need to assume $x$ is in $P$, so



      $$
      Ax geq b
      $$



      Also, $x in u+L$, i.e., $Cx=d$ where $d=Cu$.



      Now I need to show there exist an $A'$ and $b'$ where $A'x geq b'$.
      How can I proceed?







      linear-algebra convex-analysis






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      edited Dec 8 '18 at 4:49







      Saeed

















      asked Dec 8 '18 at 0:00









      SaeedSaeed

      1,024310




      1,024310






















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          $begingroup$

          Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think I got what you mean, just change the last $leq$ to $geq$.
            $endgroup$
            – Saeed
            Dec 8 '18 at 5:02











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          $begingroup$

          Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think I got what you mean, just change the last $leq$ to $geq$.
            $endgroup$
            – Saeed
            Dec 8 '18 at 5:02
















          1












          $begingroup$

          Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think I got what you mean, just change the last $leq$ to $geq$.
            $endgroup$
            – Saeed
            Dec 8 '18 at 5:02














          1












          1








          1





          $begingroup$

          Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.






          share|cite|improve this answer









          $endgroup$



          Hint: first express your affine space as a system of linear equations $Cx = d$. Then append the rows of $C$ and $-C$ to $A$ to form a taller matrix $A$, as well as append $d$ and $-d$ to $b$ to form a taller column vector $b'$. Then the inequality $A'x le b'$ describes the intersection.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 0:24









          Theo BenditTheo Bendit

          18.1k12152




          18.1k12152












          • $begingroup$
            I think I got what you mean, just change the last $leq$ to $geq$.
            $endgroup$
            – Saeed
            Dec 8 '18 at 5:02


















          • $begingroup$
            I think I got what you mean, just change the last $leq$ to $geq$.
            $endgroup$
            – Saeed
            Dec 8 '18 at 5:02
















          $begingroup$
          I think I got what you mean, just change the last $leq$ to $geq$.
          $endgroup$
          – Saeed
          Dec 8 '18 at 5:02




          $begingroup$
          I think I got what you mean, just change the last $leq$ to $geq$.
          $endgroup$
          – Saeed
          Dec 8 '18 at 5:02


















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