Projection in $ell_1$ norm onto linear subspace of $mathbb{R}^n$.












2












$begingroup$


In the context of $mathbb{R}^n$, fix a linear subspace $mathcal{U}$ or arbitrary dimension $r leq n$ and some vector $mathbf{x} in mathbb{R}^n$.



What is the most efficient way to compute the projection of $mathbf{x}$ onto $mathcal{U}$ in $ell_1$, that is, how had one ought to solve the optimization problem:



$mathbf{u}^* = min_{mathbf{u} in mathcal{U}} ||mathbf{u} - mathbf{x}||_1$



I have a hard time believing this is not a well understood problem but I am having difficulty with Google.



One could clearly formulate this as a linear program, but I am primarily concerned with high dimensional problems for $n$ large. Is there a more efficient way?






linear-algebra vector-spaces norm projection







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    In the context of $mathbb{R}^n$, fix a linear subspace $mathcal{U}$ or arbitrary dimension $r leq n$ and some vector $mathbf{x} in mathbb{R}^n$.



    What is the most efficient way to compute the projection of $mathbf{x}$ onto $mathcal{U}$ in $ell_1$, that is, how had one ought to solve the optimization problem:



    $mathbf{u}^* = min_{mathbf{u} in mathcal{U}} ||mathbf{u} - mathbf{x}||_1$



    I have a hard time believing this is not a well understood problem but I am having difficulty with Google.



    One could clearly formulate this as a linear program, but I am primarily concerned with high dimensional problems for $n$ large. Is there a more efficient way?






    linear-algebra vector-spaces norm projection







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      In the context of $mathbb{R}^n$, fix a linear subspace $mathcal{U}$ or arbitrary dimension $r leq n$ and some vector $mathbf{x} in mathbb{R}^n$.



      What is the most efficient way to compute the projection of $mathbf{x}$ onto $mathcal{U}$ in $ell_1$, that is, how had one ought to solve the optimization problem:



      $mathbf{u}^* = min_{mathbf{u} in mathcal{U}} ||mathbf{u} - mathbf{x}||_1$



      I have a hard time believing this is not a well understood problem but I am having difficulty with Google.



      One could clearly formulate this as a linear program, but I am primarily concerned with high dimensional problems for $n$ large. Is there a more efficient way?






      linear-algebra vector-spaces norm projection







      share|cite|improve this question











      $endgroup$




      In the context of $mathbb{R}^n$, fix a linear subspace $mathcal{U}$ or arbitrary dimension $r leq n$ and some vector $mathbf{x} in mathbb{R}^n$.



      What is the most efficient way to compute the projection of $mathbf{x}$ onto $mathcal{U}$ in $ell_1$, that is, how had one ought to solve the optimization problem:



      $mathbf{u}^* = min_{mathbf{u} in mathcal{U}} ||mathbf{u} - mathbf{x}||_1$



      I have a hard time believing this is not a well understood problem but I am having difficulty with Google.



      One could clearly formulate this as a linear program, but I am primarily concerned with high dimensional problems for $n$ large. Is there a more efficient way?






      linear-algebra vector-spaces norm projection




      linear-algebra vector-spaces norm projection






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 15:13







      John Madden

















      asked Dec 27 '18 at 15:00









      John MaddenJohn Madden

      1676




      1676






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Well, $displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=min{rinmathbb{R}_{geqslant 0} : mathcal{U}cap B_1(mathbf{x},r)neqemptyset}$, where
          $$B_1(mathbf{x},r)={mathbf{y}inmathbb{R}^n : ||mathbf{y}-mathbf{x}||_1leqslant r}$$
          is the closed $l_1$-ball with center $mathbf{x}$ and radius $r$. It is the convex hull of
          $$E_1(mathbf{x},r)={mathbf{x}pm rmathbf{e}_k : 1 leqslant k leqslant n},$$
          where $(mathbf{e}_1,ldots,mathbf{e}_n)$ is the standard basis of $mathbb{R}^n$. It follows that
          $$displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=r^astimpliesmathcal{U}cap E_1(mathbf{x},r^ast)neqemptyset,$$
          the minimum is attained at any (not necessarily unique) element $mathbf{u}^ast$ of the last set, and the convex hull of all such $mathbf{u}^ast$ is the solution to the original problem.



          So the whole problem reduces to finding the minimum of at most $n$ real numbers
          $$r_k = inf{|r| : mathbf{x}+rmathbf{e}_kinmathcal{U}},quad 1leqslant kleqslant n$$
          (with $inf emptyset$ considered to be $+infty$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            (The same with arbitrary norms - the unit sphere being crucial to consider.)
            $endgroup$
            – metamorphy
            Dec 27 '18 at 18:41










          • $begingroup$
            I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:37










          • $begingroup$
            Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:38










          • $begingroup$
            Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
            $endgroup$
            – metamorphy
            Dec 28 '18 at 16:11











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054018%2fprojection-in-ell-1-norm-onto-linear-subspace-of-mathbbrn%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Well, $displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=min{rinmathbb{R}_{geqslant 0} : mathcal{U}cap B_1(mathbf{x},r)neqemptyset}$, where
          $$B_1(mathbf{x},r)={mathbf{y}inmathbb{R}^n : ||mathbf{y}-mathbf{x}||_1leqslant r}$$
          is the closed $l_1$-ball with center $mathbf{x}$ and radius $r$. It is the convex hull of
          $$E_1(mathbf{x},r)={mathbf{x}pm rmathbf{e}_k : 1 leqslant k leqslant n},$$
          where $(mathbf{e}_1,ldots,mathbf{e}_n)$ is the standard basis of $mathbb{R}^n$. It follows that
          $$displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=r^astimpliesmathcal{U}cap E_1(mathbf{x},r^ast)neqemptyset,$$
          the minimum is attained at any (not necessarily unique) element $mathbf{u}^ast$ of the last set, and the convex hull of all such $mathbf{u}^ast$ is the solution to the original problem.



          So the whole problem reduces to finding the minimum of at most $n$ real numbers
          $$r_k = inf{|r| : mathbf{x}+rmathbf{e}_kinmathcal{U}},quad 1leqslant kleqslant n$$
          (with $inf emptyset$ considered to be $+infty$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            (The same with arbitrary norms - the unit sphere being crucial to consider.)
            $endgroup$
            – metamorphy
            Dec 27 '18 at 18:41










          • $begingroup$
            I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:37










          • $begingroup$
            Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:38










          • $begingroup$
            Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
            $endgroup$
            – metamorphy
            Dec 28 '18 at 16:11
















          1












          $begingroup$

          Well, $displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=min{rinmathbb{R}_{geqslant 0} : mathcal{U}cap B_1(mathbf{x},r)neqemptyset}$, where
          $$B_1(mathbf{x},r)={mathbf{y}inmathbb{R}^n : ||mathbf{y}-mathbf{x}||_1leqslant r}$$
          is the closed $l_1$-ball with center $mathbf{x}$ and radius $r$. It is the convex hull of
          $$E_1(mathbf{x},r)={mathbf{x}pm rmathbf{e}_k : 1 leqslant k leqslant n},$$
          where $(mathbf{e}_1,ldots,mathbf{e}_n)$ is the standard basis of $mathbb{R}^n$. It follows that
          $$displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=r^astimpliesmathcal{U}cap E_1(mathbf{x},r^ast)neqemptyset,$$
          the minimum is attained at any (not necessarily unique) element $mathbf{u}^ast$ of the last set, and the convex hull of all such $mathbf{u}^ast$ is the solution to the original problem.



          So the whole problem reduces to finding the minimum of at most $n$ real numbers
          $$r_k = inf{|r| : mathbf{x}+rmathbf{e}_kinmathcal{U}},quad 1leqslant kleqslant n$$
          (with $inf emptyset$ considered to be $+infty$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            (The same with arbitrary norms - the unit sphere being crucial to consider.)
            $endgroup$
            – metamorphy
            Dec 27 '18 at 18:41










          • $begingroup$
            I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:37










          • $begingroup$
            Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:38










          • $begingroup$
            Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
            $endgroup$
            – metamorphy
            Dec 28 '18 at 16:11














          1












          1








          1





          $begingroup$

          Well, $displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=min{rinmathbb{R}_{geqslant 0} : mathcal{U}cap B_1(mathbf{x},r)neqemptyset}$, where
          $$B_1(mathbf{x},r)={mathbf{y}inmathbb{R}^n : ||mathbf{y}-mathbf{x}||_1leqslant r}$$
          is the closed $l_1$-ball with center $mathbf{x}$ and radius $r$. It is the convex hull of
          $$E_1(mathbf{x},r)={mathbf{x}pm rmathbf{e}_k : 1 leqslant k leqslant n},$$
          where $(mathbf{e}_1,ldots,mathbf{e}_n)$ is the standard basis of $mathbb{R}^n$. It follows that
          $$displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=r^astimpliesmathcal{U}cap E_1(mathbf{x},r^ast)neqemptyset,$$
          the minimum is attained at any (not necessarily unique) element $mathbf{u}^ast$ of the last set, and the convex hull of all such $mathbf{u}^ast$ is the solution to the original problem.



          So the whole problem reduces to finding the minimum of at most $n$ real numbers
          $$r_k = inf{|r| : mathbf{x}+rmathbf{e}_kinmathcal{U}},quad 1leqslant kleqslant n$$
          (with $inf emptyset$ considered to be $+infty$).






          share|cite|improve this answer











          $endgroup$



          Well, $displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=min{rinmathbb{R}_{geqslant 0} : mathcal{U}cap B_1(mathbf{x},r)neqemptyset}$, where
          $$B_1(mathbf{x},r)={mathbf{y}inmathbb{R}^n : ||mathbf{y}-mathbf{x}||_1leqslant r}$$
          is the closed $l_1$-ball with center $mathbf{x}$ and radius $r$. It is the convex hull of
          $$E_1(mathbf{x},r)={mathbf{x}pm rmathbf{e}_k : 1 leqslant k leqslant n},$$
          where $(mathbf{e}_1,ldots,mathbf{e}_n)$ is the standard basis of $mathbb{R}^n$. It follows that
          $$displaystylemin_{mathbf{u}inmathcal{U}}||mathbf{u}-mathbf{x}||_1=r^astimpliesmathcal{U}cap E_1(mathbf{x},r^ast)neqemptyset,$$
          the minimum is attained at any (not necessarily unique) element $mathbf{u}^ast$ of the last set, and the convex hull of all such $mathbf{u}^ast$ is the solution to the original problem.



          So the whole problem reduces to finding the minimum of at most $n$ real numbers
          $$r_k = inf{|r| : mathbf{x}+rmathbf{e}_kinmathcal{U}},quad 1leqslant kleqslant n$$
          (with $inf emptyset$ considered to be $+infty$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 28 '18 at 8:26

























          answered Dec 27 '18 at 17:42









          metamorphymetamorphy

          3,6821621




          3,6821621












          • $begingroup$
            (The same with arbitrary norms - the unit sphere being crucial to consider.)
            $endgroup$
            – metamorphy
            Dec 27 '18 at 18:41










          • $begingroup$
            I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:37










          • $begingroup$
            Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:38










          • $begingroup$
            Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
            $endgroup$
            – metamorphy
            Dec 28 '18 at 16:11


















          • $begingroup$
            (The same with arbitrary norms - the unit sphere being crucial to consider.)
            $endgroup$
            – metamorphy
            Dec 27 '18 at 18:41










          • $begingroup$
            I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:37










          • $begingroup$
            Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
            $endgroup$
            – John Madden
            Dec 27 '18 at 21:38










          • $begingroup$
            Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
            $endgroup$
            – metamorphy
            Dec 28 '18 at 16:11
















          $begingroup$
          (The same with arbitrary norms - the unit sphere being crucial to consider.)
          $endgroup$
          – metamorphy
          Dec 27 '18 at 18:41




          $begingroup$
          (The same with arbitrary norms - the unit sphere being crucial to consider.)
          $endgroup$
          – metamorphy
          Dec 27 '18 at 18:41












          $begingroup$
          I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
          $endgroup$
          – John Madden
          Dec 27 '18 at 21:37




          $begingroup$
          I see, so projection in $ell_1$ generally only involves changing one coordinate of the vector to be projected? Perhaps this explains why it is not often used. Interesting, I imagine it is easy to compute the infimum in your final expression?
          $endgroup$
          – John Madden
          Dec 27 '18 at 21:37












          $begingroup$
          Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
          $endgroup$
          – John Madden
          Dec 27 '18 at 21:38




          $begingroup$
          Also, it seems that my use of a double negative made the expression difficult to understand: what I'm effectively saying is: "surely this problem is well understood' in my original post.
          $endgroup$
          – John Madden
          Dec 27 '18 at 21:38












          $begingroup$
          Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
          $endgroup$
          – metamorphy
          Dec 28 '18 at 16:11




          $begingroup$
          Yes, the infimum is easy to compute (either it is $0$, or there is at most single $r$). For "double negative", I'm sorry for inattentive reading ;)
          $endgroup$
          – metamorphy
          Dec 28 '18 at 16:11


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054018%2fprojection-in-ell-1-norm-onto-linear-subspace-of-mathbbrn%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more