Bilinear form in finite dimension space












0












$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38
















0












$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38














0












0








0





$begingroup$


How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra







share|cite|improve this question











$endgroup$




How to prove that there exist a unique operator $T:Vto V$ such that



$$f(x,y)=g(Tx,y)$$



where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.



I proved the uniqueness but I dont know how to prove the existence.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:39







Bilal Jafar Karaki

















asked Dec 21 '18 at 17:31









Bilal Jafar KarakiBilal Jafar Karaki

392113




392113








  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38














  • 2




    $begingroup$
    We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
    $endgroup$
    – Omnomnomnom
    Dec 21 '18 at 17:37










  • $begingroup$
    yes, g is a non-degenerate bilinear form.
    $endgroup$
    – Bilal Jafar Karaki
    Dec 21 '18 at 17:38








2




2




$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37




$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37












$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38




$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
    $$
    hat{g}(x)colon ymapsto g(x,y)
    $$
    Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



    Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
    $$
    hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
    $$
    and therefore, for every $yin V$,
    $$
    hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
    $$
    which is the same as saying that $f(x,y)=g(Tx,y)$.




    How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






    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%2f3048728%2fbilinear-form-in-finite-dimension-space%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









      1












      $begingroup$

      Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
      $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
      Now show there exists a matrix $T$ such that $T^top B = A$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
        $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
        Now show there exists a matrix $T$ such that $T^top B = A$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
          $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
          Now show there exists a matrix $T$ such that $T^top B = A$.






          share|cite|improve this answer









          $endgroup$



          Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
          $$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
          Now show there exists a matrix $T$ such that $T^top B = A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 17:40









          SvanNSvanN

          2,0661422




          2,0661422























              1












              $begingroup$

              The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
              $$
              hat{g}(x)colon ymapsto g(x,y)
              $$
              Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



              Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
              $$
              hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
              $$
              and therefore, for every $yin V$,
              $$
              hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
              $$
              which is the same as saying that $f(x,y)=g(Tx,y)$.




              How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                $$
                hat{g}(x)colon ymapsto g(x,y)
                $$
                Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                $$
                hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                $$
                and therefore, for every $yin V$,
                $$
                hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                $$
                which is the same as saying that $f(x,y)=g(Tx,y)$.




                How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                  $$
                  hat{g}(x)colon ymapsto g(x,y)
                  $$
                  Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                  Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                  $$
                  hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                  $$
                  and therefore, for every $yin V$,
                  $$
                  hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                  $$
                  which is the same as saying that $f(x,y)=g(Tx,y)$.




                  How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.






                  share|cite|improve this answer









                  $endgroup$



                  The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
                  $$
                  hat{g}(x)colon ymapsto g(x,y)
                  $$
                  Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.



                  Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
                  $$
                  hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
                  $$
                  and therefore, for every $yin V$,
                  $$
                  hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
                  $$
                  which is the same as saying that $f(x,y)=g(Tx,y)$.




                  How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 18:47









                  egregegreg

                  185k1486206




                  185k1486206






























                      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%2f3048728%2fbilinear-form-in-finite-dimension-space%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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more