Eigenspaces of an orthogonal projection











up vote
1
down vote

favorite












So for an orthogonal projection $P:Vrightarrow U $, the task is to find the eigenvalues and eigenspaces of P.



I have found that $lambda = 0,1$



Then $E_0 = ker(P) = U^{perp}$



but for $E_1$ i'm not sure



$E_1 = ker(P-Id)$



I feel like the solution for this is somewhat trivial and is staring me in the face, but i cant see it



Can anyone explain? thanks






eigenvalues-eigenvectors







share|cite|improve this question


















  • 1




    What actually do you want to find?
    – Anupam
    Nov 22 at 4:30















up vote
1
down vote

favorite












So for an orthogonal projection $P:Vrightarrow U $, the task is to find the eigenvalues and eigenspaces of P.



I have found that $lambda = 0,1$



Then $E_0 = ker(P) = U^{perp}$



but for $E_1$ i'm not sure



$E_1 = ker(P-Id)$



I feel like the solution for this is somewhat trivial and is staring me in the face, but i cant see it



Can anyone explain? thanks






eigenvalues-eigenvectors







share|cite|improve this question


















  • 1




    What actually do you want to find?
    – Anupam
    Nov 22 at 4:30













up vote
1
down vote

favorite









up vote
1
down vote

favorite











So for an orthogonal projection $P:Vrightarrow U $, the task is to find the eigenvalues and eigenspaces of P.



I have found that $lambda = 0,1$



Then $E_0 = ker(P) = U^{perp}$



but for $E_1$ i'm not sure



$E_1 = ker(P-Id)$



I feel like the solution for this is somewhat trivial and is staring me in the face, but i cant see it



Can anyone explain? thanks






eigenvalues-eigenvectors







share|cite|improve this question













So for an orthogonal projection $P:Vrightarrow U $, the task is to find the eigenvalues and eigenspaces of P.



I have found that $lambda = 0,1$



Then $E_0 = ker(P) = U^{perp}$



but for $E_1$ i'm not sure



$E_1 = ker(P-Id)$



I feel like the solution for this is somewhat trivial and is staring me in the face, but i cant see it



Can anyone explain? thanks






eigenvalues-eigenvectors




eigenvalues-eigenvectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 4:20









big daddy

434




434








  • 1




    What actually do you want to find?
    – Anupam
    Nov 22 at 4:30














  • 1




    What actually do you want to find?
    – Anupam
    Nov 22 at 4:30








1




1




What actually do you want to find?
– Anupam
Nov 22 at 4:30




What actually do you want to find?
– Anupam
Nov 22 at 4:30










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Let $x in E_1$. Then, $x=P(x) in P(V)$. Hence, $E_1 subseteq P(V)$.



Let, $y in P(V)$. Then, $y=P(X)$ for some $x in V$. Hence, $P(y)=P^2(x)=P(x)=y$ i.e. $y in E_1$. $therefore P(V) subseteq E_1$.



Combining, we get, $P(V)=E_1$.






share|cite|improve this answer




























    up vote
    1
    down vote













    $E_1$ can be proved to be the range of P. You'll have to do a bit of work, but it's not too hard to show.






    share|cite|improve this answer





















      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',
      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%2f3008736%2feigenspaces-of-an-orthogonal-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








      up vote
      1
      down vote



      accepted










      Let $x in E_1$. Then, $x=P(x) in P(V)$. Hence, $E_1 subseteq P(V)$.



      Let, $y in P(V)$. Then, $y=P(X)$ for some $x in V$. Hence, $P(y)=P^2(x)=P(x)=y$ i.e. $y in E_1$. $therefore P(V) subseteq E_1$.



      Combining, we get, $P(V)=E_1$.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Let $x in E_1$. Then, $x=P(x) in P(V)$. Hence, $E_1 subseteq P(V)$.



        Let, $y in P(V)$. Then, $y=P(X)$ for some $x in V$. Hence, $P(y)=P^2(x)=P(x)=y$ i.e. $y in E_1$. $therefore P(V) subseteq E_1$.



        Combining, we get, $P(V)=E_1$.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Let $x in E_1$. Then, $x=P(x) in P(V)$. Hence, $E_1 subseteq P(V)$.



          Let, $y in P(V)$. Then, $y=P(X)$ for some $x in V$. Hence, $P(y)=P^2(x)=P(x)=y$ i.e. $y in E_1$. $therefore P(V) subseteq E_1$.



          Combining, we get, $P(V)=E_1$.






          share|cite|improve this answer












          Let $x in E_1$. Then, $x=P(x) in P(V)$. Hence, $E_1 subseteq P(V)$.



          Let, $y in P(V)$. Then, $y=P(X)$ for some $x in V$. Hence, $P(y)=P^2(x)=P(x)=y$ i.e. $y in E_1$. $therefore P(V) subseteq E_1$.



          Combining, we get, $P(V)=E_1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 5:51









          SinTan1729

          2,399622




          2,399622






















              up vote
              1
              down vote













              $E_1$ can be proved to be the range of P. You'll have to do a bit of work, but it's not too hard to show.






              share|cite|improve this answer

























                up vote
                1
                down vote













                $E_1$ can be proved to be the range of P. You'll have to do a bit of work, but it's not too hard to show.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $E_1$ can be proved to be the range of P. You'll have to do a bit of work, but it's not too hard to show.






                  share|cite|improve this answer












                  $E_1$ can be proved to be the range of P. You'll have to do a bit of work, but it's not too hard to show.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 5:05









                  JonathanZ

                  2,089613




                  2,089613






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3008736%2feigenspaces-of-an-orthogonal-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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more