If a vector is orthogonal to a subspace, what is its projection?












0












$begingroup$


Suppose $y$ is orthogonal to a subspace spanned by the columns of matrix $A$.



My question is, what is the projection of $y$ onto $A$?



I know that the projection of $y$ onto $A$ is the vector in $A$ such that the distance between that vector and $y$ is minimal.



But I think every vector in $A$ in this case attains same distance.



Hence the projection of $y$ onto $A$ is the set of vectors spanned by $A$.



Is my intuition correct?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
    $endgroup$
    – John Hughes
    Jan 23 at 8:08


















0












$begingroup$


Suppose $y$ is orthogonal to a subspace spanned by the columns of matrix $A$.



My question is, what is the projection of $y$ onto $A$?



I know that the projection of $y$ onto $A$ is the vector in $A$ such that the distance between that vector and $y$ is minimal.



But I think every vector in $A$ in this case attains same distance.



Hence the projection of $y$ onto $A$ is the set of vectors spanned by $A$.



Is my intuition correct?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
    $endgroup$
    – John Hughes
    Jan 23 at 8:08
















0












0








0





$begingroup$


Suppose $y$ is orthogonal to a subspace spanned by the columns of matrix $A$.



My question is, what is the projection of $y$ onto $A$?



I know that the projection of $y$ onto $A$ is the vector in $A$ such that the distance between that vector and $y$ is minimal.



But I think every vector in $A$ in this case attains same distance.



Hence the projection of $y$ onto $A$ is the set of vectors spanned by $A$.



Is my intuition correct?










share|cite|improve this question









$endgroup$




Suppose $y$ is orthogonal to a subspace spanned by the columns of matrix $A$.



My question is, what is the projection of $y$ onto $A$?



I know that the projection of $y$ onto $A$ is the vector in $A$ such that the distance between that vector and $y$ is minimal.



But I think every vector in $A$ in this case attains same distance.



Hence the projection of $y$ onto $A$ is the set of vectors spanned by $A$.



Is my intuition correct?







linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 at 6:37









The man of your dreamThe man of your dream

23110




23110












  • $begingroup$
    Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
    $endgroup$
    – John Hughes
    Jan 23 at 8:08




















  • $begingroup$
    Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
    $endgroup$
    – John Hughes
    Jan 23 at 8:08


















$begingroup$
Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
$endgroup$
– John Hughes
Jan 23 at 8:08






$begingroup$
Your question, and both answers given so far, assumes that there is a projection of $y$ onto the subspace $S_A$ spanned by the columns of $A$. But consider $pmatrix{0 & 0 \ 1 & 2}$, and $y = pmatrix{1\2}$. The subspace $S_A$ is the $y$-axis, so you might say that "the projection" of $y$ is $pmatrix{0\2}$. But the transformation represented by $B = pmatrix{0 &0 \ 1 & 1}$ is also a projection onto $S_A$, and takes $y$ to $pmatrix{0\3}$. It just happens to not be an orthogonal projection.
$endgroup$
– John Hughes
Jan 23 at 8:08












2 Answers
2






active

oldest

votes


















0












$begingroup$

The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Zero.



    The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{perp}$.



    That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.






    share|cite|improve this answer











    $endgroup$













      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%2f3084156%2fif-a-vector-is-orthogonal-to-a-subspace-what-is-its-projection%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$






          share|cite|improve this answer









          $endgroup$



          The projection of $y$ onto $A$ will be the zero vector. For any other vector $vin A$ you will have $||y-v|| = sqrt{||y||^2+||v||^2} > ||y||.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 6:42









          Reiner MartinReiner Martin

          3,434414




          3,434414























              0












              $begingroup$

              Zero.



              The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{perp}$.



              That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Zero.



                The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{perp}$.



                That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Zero.



                  The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{perp}$.



                  That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.






                  share|cite|improve this answer











                  $endgroup$



                  Zero.



                  The kernel of the projection onto a subspace $A$ is the orthogonal complement of the subspace, $A^{perp}$.



                  That should be your intuition. The zero vector is in a sense the closest vector in $A$ to an orthogonal vector.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 7:47

























                  answered Jan 23 at 7:42









                  Chris CusterChris Custer

                  14k3827




                  14k3827






























                      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%2f3084156%2fif-a-vector-is-orthogonal-to-a-subspace-what-is-its-projection%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

                      Mario Kart Wii

                      What does “Dominus providebit” mean?

                      Antonio Litta Visconti Arese