prove Gram's determinant invariant under orthogonalisation











up vote
0
down vote

favorite












let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



$left|{begin{array}{ccc}
left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
vdots & ddots & vdots\
left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
end{array}}right|$



write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
$G(g_1,...g_n)=G(h_1,...h_n)$



It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






functional-analysis hilbert-spaces







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



    $left|{begin{array}{ccc}
    left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
    vdots & ddots & vdots\
    left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
    end{array}}right|$



    write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
    $G(g_1,...g_n)=G(h_1,...h_n)$



    It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






    functional-analysis hilbert-spaces







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



      $left|{begin{array}{ccc}
      left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
      vdots & ddots & vdots\
      left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
      end{array}}right|$



      write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
      $G(g_1,...g_n)=G(h_1,...h_n)$



      It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






      functional-analysis hilbert-spaces







      share|cite|improve this question















      let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:



      $left|{begin{array}{ccc}
      left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
      vdots & ddots & vdots\
      left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
      end{array}}right|$



      write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
      $G(g_1,...g_n)=G(h_1,...h_n)$



      It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.






      functional-analysis hilbert-spaces




      functional-analysis hilbert-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 15:57

























      asked Nov 22 at 14:46









      S. R

      886




      886






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote



          accepted










          Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
          $$
          mathcal G(g_1,h_2)=
          begin{bmatrix}
          |g_1|^2 & (g_1,h_2)\
          (h_2,g_1) & |g_2|^2
          end{bmatrix}
          =
          begin{bmatrix}
          |g_1|^2 & t-t|g_1|^2\
          t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
          end{bmatrix}
          $$
          This matrix can be obtained from
          $$ mathcal G (g_1,g_2) =
          begin{bmatrix}
          |g_1|^2 & t\
          t & |g_2|^2
          end{bmatrix}
          $$
          by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



          For higher dimensions you see the emergence of terms
          begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
          (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
          The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
          $$ (g_2,g_i) - t(g_1,g_i)$$
          as needed, except the diagonal term.
          If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






          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%2f3009223%2fprove-grams-determinant-invariant-under-orthogonalisation%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








            up vote
            0
            down vote



            accepted










            Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
            $$
            mathcal G(g_1,h_2)=
            begin{bmatrix}
            |g_1|^2 & (g_1,h_2)\
            (h_2,g_1) & |g_2|^2
            end{bmatrix}
            =
            begin{bmatrix}
            |g_1|^2 & t-t|g_1|^2\
            t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
            end{bmatrix}
            $$
            This matrix can be obtained from
            $$ mathcal G (g_1,g_2) =
            begin{bmatrix}
            |g_1|^2 & t\
            t & |g_2|^2
            end{bmatrix}
            $$
            by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



            For higher dimensions you see the emergence of terms
            begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
            (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
            The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
            $$ (g_2,g_i) - t(g_1,g_i)$$
            as needed, except the diagonal term.
            If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






            share|cite|improve this answer



























              up vote
              0
              down vote



              accepted










              Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
              $$
              mathcal G(g_1,h_2)=
              begin{bmatrix}
              |g_1|^2 & (g_1,h_2)\
              (h_2,g_1) & |g_2|^2
              end{bmatrix}
              =
              begin{bmatrix}
              |g_1|^2 & t-t|g_1|^2\
              t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
              end{bmatrix}
              $$
              This matrix can be obtained from
              $$ mathcal G (g_1,g_2) =
              begin{bmatrix}
              |g_1|^2 & t\
              t & |g_2|^2
              end{bmatrix}
              $$
              by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



              For higher dimensions you see the emergence of terms
              begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
              (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
              The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
              $$ (g_2,g_i) - t(g_1,g_i)$$
              as needed, except the diagonal term.
              If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






              share|cite|improve this answer

























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
                $$
                mathcal G(g_1,h_2)=
                begin{bmatrix}
                |g_1|^2 & (g_1,h_2)\
                (h_2,g_1) & |g_2|^2
                end{bmatrix}
                =
                begin{bmatrix}
                |g_1|^2 & t-t|g_1|^2\
                t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
                end{bmatrix}
                $$
                This matrix can be obtained from
                $$ mathcal G (g_1,g_2) =
                begin{bmatrix}
                |g_1|^2 & t\
                t & |g_2|^2
                end{bmatrix}
                $$
                by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



                For higher dimensions you see the emergence of terms
                begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
                (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
                The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
                $$ (g_2,g_i) - t(g_1,g_i)$$
                as needed, except the diagonal term.
                If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.






                share|cite|improve this answer














                Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
                $$
                mathcal G(g_1,h_2)=
                begin{bmatrix}
                |g_1|^2 & (g_1,h_2)\
                (h_2,g_1) & |g_2|^2
                end{bmatrix}
                =
                begin{bmatrix}
                |g_1|^2 & t-t|g_1|^2\
                t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
                end{bmatrix}
                $$
                This matrix can be obtained from
                $$ mathcal G (g_1,g_2) =
                begin{bmatrix}
                |g_1|^2 & t\
                t & |g_2|^2
                end{bmatrix}
                $$
                by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.



                For higher dimensions you see the emergence of terms
                begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
                (h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
                The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
                $$ (g_2,g_i) - t(g_1,g_i)$$
                as needed, except the diagonal term.
                If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 22 at 17:40

























                answered Nov 22 at 17:27









                Calvin Khor

                11.2k21438




                11.2k21438






























                    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%2f3009223%2fprove-grams-determinant-invariant-under-orthogonalisation%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