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












2














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







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    2














    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



























      2












      2








      2







      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















      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

























      asked Dec 27 '18 at 15:00









      John Madden

      1676




      1676






















          1 Answer
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          1














          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























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










          • 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?
            – John Madden
            Dec 27 '18 at 21:37










          • 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.
            – John Madden
            Dec 27 '18 at 21:38










          • 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 ;)
            – metamorphy
            Dec 28 '18 at 16:11











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          1 Answer
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          active

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          active

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          1














          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























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










          • 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?
            – John Madden
            Dec 27 '18 at 21:37










          • 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.
            – John Madden
            Dec 27 '18 at 21:38










          • 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 ;)
            – metamorphy
            Dec 28 '18 at 16:11
















          1














          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























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










          • 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?
            – John Madden
            Dec 27 '18 at 21:37










          • 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.
            – John Madden
            Dec 27 '18 at 21:38










          • 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 ;)
            – metamorphy
            Dec 28 '18 at 16:11














          1












          1








          1






          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














          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









          metamorphy

          3,5571521




          3,5571521












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










          • 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?
            – John Madden
            Dec 27 '18 at 21:37










          • 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.
            – John Madden
            Dec 27 '18 at 21:38










          • 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 ;)
            – metamorphy
            Dec 28 '18 at 16:11


















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










          • 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?
            – John Madden
            Dec 27 '18 at 21:37










          • 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.
            – John Madden
            Dec 27 '18 at 21:38










          • 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 ;)
            – metamorphy
            Dec 28 '18 at 16:11
















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




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












          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?
          – John Madden
          Dec 27 '18 at 21:37




          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?
          – John Madden
          Dec 27 '18 at 21:37












          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.
          – John Madden
          Dec 27 '18 at 21:38




          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.
          – John Madden
          Dec 27 '18 at 21:38












          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 ;)
          – metamorphy
          Dec 28 '18 at 16:11




          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 ;)
          – metamorphy
          Dec 28 '18 at 16:11


















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