Identity between resolvent and singular value density












0














I was reading the paper




Sengupta, Anirvan M., and Partha P. Mitra. "Distributions of singular
values for some random matrices." Physical Review E 60.3 (1999): 3389.




but got stuck at equation (3):



enter image description here



Here $lambda_n$ are the singular values of a matrix $M$ (eventually taken to be a random matrix), $rho(lambda)$ denotes the density of singular values of $M$, and the resolvent $mathcal G(z)$ is defined by:



enter image description here



So how does (3) follow from (2)?






linear-algebra random-matrices singularvalues







share|cite|improve this question



























    0














    I was reading the paper




    Sengupta, Anirvan M., and Partha P. Mitra. "Distributions of singular
    values for some random matrices." Physical Review E 60.3 (1999): 3389.




    but got stuck at equation (3):



    enter image description here



    Here $lambda_n$ are the singular values of a matrix $M$ (eventually taken to be a random matrix), $rho(lambda)$ denotes the density of singular values of $M$, and the resolvent $mathcal G(z)$ is defined by:



    enter image description here



    So how does (3) follow from (2)?






    linear-algebra random-matrices singularvalues







    share|cite|improve this question

























      0












      0








      0







      I was reading the paper




      Sengupta, Anirvan M., and Partha P. Mitra. "Distributions of singular
      values for some random matrices." Physical Review E 60.3 (1999): 3389.




      but got stuck at equation (3):



      enter image description here



      Here $lambda_n$ are the singular values of a matrix $M$ (eventually taken to be a random matrix), $rho(lambda)$ denotes the density of singular values of $M$, and the resolvent $mathcal G(z)$ is defined by:



      enter image description here



      So how does (3) follow from (2)?






      linear-algebra random-matrices singularvalues







      share|cite|improve this question













      I was reading the paper




      Sengupta, Anirvan M., and Partha P. Mitra. "Distributions of singular
      values for some random matrices." Physical Review E 60.3 (1999): 3389.




      but got stuck at equation (3):



      enter image description here



      Here $lambda_n$ are the singular values of a matrix $M$ (eventually taken to be a random matrix), $rho(lambda)$ denotes the density of singular values of $M$, and the resolvent $mathcal G(z)$ is defined by:



      enter image description here



      So how does (3) follow from (2)?






      linear-algebra random-matrices singularvalues




      linear-algebra random-matrices singularvalues






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 21:19









      beckobecko

      2,34931942




      2,34931942






















          1 Answer
          1






          active

          oldest

          votes


















          1














          I also have problems, but let me write what I got:



          Perhaps you know that $lim_{epsilonto 0^+}frac{1}{pi}Im(frac{1}{t-iepsilon})=delta(t)$. In any case, this follows from $Im(frac{1}{t-iepsilon})=frac{1}{epsilon((t/epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2lambdasum_ndelta(lambda^2-lambda_n^2)$, so we should prove that $2tdelta(t^2-a^2)=delta(t-a)$. Unfortunately, I prove instead that $2tdelta(t^2-a^2)=-delta(t+|a|)+delta(t-|a|)$.



          It's easier to approximate $delta$ by $phi_epsilon=epsilon^{-1}phi(t/epsilon)$, where $phi$ is compactly supported and even. Now, let $psi$ be a test function and evaluate
          $$
          begin{align}
          int 2tphi_epsilon(t^2-a^2)psi(t),dt &= int_{-infty}^0+int_0^{infty} \
          &=-int_{-a^2}^infty phi_epsilon(s)psi(-sqrt{s+a^2}),ds +int_{-a^2}^inftyphi_epsilon(s)psi(sqrt{s+a^2}),ds,
          end{align}
          $$
          where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $epsilonto 0$ we get $2tdelta(t^2-a^2) = -delta(t+|a|)+ delta(t-|a|)$. I don't know if I made a mistake.






          share|cite|improve this answer

















          • 1




            I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
            – becko
            Nov 30 '18 at 20:32













          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%2f3019236%2fidentity-between-resolvent-and-singular-value-density%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









          1














          I also have problems, but let me write what I got:



          Perhaps you know that $lim_{epsilonto 0^+}frac{1}{pi}Im(frac{1}{t-iepsilon})=delta(t)$. In any case, this follows from $Im(frac{1}{t-iepsilon})=frac{1}{epsilon((t/epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2lambdasum_ndelta(lambda^2-lambda_n^2)$, so we should prove that $2tdelta(t^2-a^2)=delta(t-a)$. Unfortunately, I prove instead that $2tdelta(t^2-a^2)=-delta(t+|a|)+delta(t-|a|)$.



          It's easier to approximate $delta$ by $phi_epsilon=epsilon^{-1}phi(t/epsilon)$, where $phi$ is compactly supported and even. Now, let $psi$ be a test function and evaluate
          $$
          begin{align}
          int 2tphi_epsilon(t^2-a^2)psi(t),dt &= int_{-infty}^0+int_0^{infty} \
          &=-int_{-a^2}^infty phi_epsilon(s)psi(-sqrt{s+a^2}),ds +int_{-a^2}^inftyphi_epsilon(s)psi(sqrt{s+a^2}),ds,
          end{align}
          $$
          where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $epsilonto 0$ we get $2tdelta(t^2-a^2) = -delta(t+|a|)+ delta(t-|a|)$. I don't know if I made a mistake.






          share|cite|improve this answer

















          • 1




            I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
            – becko
            Nov 30 '18 at 20:32


















          1














          I also have problems, but let me write what I got:



          Perhaps you know that $lim_{epsilonto 0^+}frac{1}{pi}Im(frac{1}{t-iepsilon})=delta(t)$. In any case, this follows from $Im(frac{1}{t-iepsilon})=frac{1}{epsilon((t/epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2lambdasum_ndelta(lambda^2-lambda_n^2)$, so we should prove that $2tdelta(t^2-a^2)=delta(t-a)$. Unfortunately, I prove instead that $2tdelta(t^2-a^2)=-delta(t+|a|)+delta(t-|a|)$.



          It's easier to approximate $delta$ by $phi_epsilon=epsilon^{-1}phi(t/epsilon)$, where $phi$ is compactly supported and even. Now, let $psi$ be a test function and evaluate
          $$
          begin{align}
          int 2tphi_epsilon(t^2-a^2)psi(t),dt &= int_{-infty}^0+int_0^{infty} \
          &=-int_{-a^2}^infty phi_epsilon(s)psi(-sqrt{s+a^2}),ds +int_{-a^2}^inftyphi_epsilon(s)psi(sqrt{s+a^2}),ds,
          end{align}
          $$
          where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $epsilonto 0$ we get $2tdelta(t^2-a^2) = -delta(t+|a|)+ delta(t-|a|)$. I don't know if I made a mistake.






          share|cite|improve this answer

















          • 1




            I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
            – becko
            Nov 30 '18 at 20:32
















          1












          1








          1






          I also have problems, but let me write what I got:



          Perhaps you know that $lim_{epsilonto 0^+}frac{1}{pi}Im(frac{1}{t-iepsilon})=delta(t)$. In any case, this follows from $Im(frac{1}{t-iepsilon})=frac{1}{epsilon((t/epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2lambdasum_ndelta(lambda^2-lambda_n^2)$, so we should prove that $2tdelta(t^2-a^2)=delta(t-a)$. Unfortunately, I prove instead that $2tdelta(t^2-a^2)=-delta(t+|a|)+delta(t-|a|)$.



          It's easier to approximate $delta$ by $phi_epsilon=epsilon^{-1}phi(t/epsilon)$, where $phi$ is compactly supported and even. Now, let $psi$ be a test function and evaluate
          $$
          begin{align}
          int 2tphi_epsilon(t^2-a^2)psi(t),dt &= int_{-infty}^0+int_0^{infty} \
          &=-int_{-a^2}^infty phi_epsilon(s)psi(-sqrt{s+a^2}),ds +int_{-a^2}^inftyphi_epsilon(s)psi(sqrt{s+a^2}),ds,
          end{align}
          $$
          where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $epsilonto 0$ we get $2tdelta(t^2-a^2) = -delta(t+|a|)+ delta(t-|a|)$. I don't know if I made a mistake.






          share|cite|improve this answer












          I also have problems, but let me write what I got:



          Perhaps you know that $lim_{epsilonto 0^+}frac{1}{pi}Im(frac{1}{t-iepsilon})=delta(t)$. In any case, this follows from $Im(frac{1}{t-iepsilon})=frac{1}{epsilon((t/epsilon)^2+1)}$. Hence, we get the right part of (3) is equal to $2lambdasum_ndelta(lambda^2-lambda_n^2)$, so we should prove that $2tdelta(t^2-a^2)=delta(t-a)$. Unfortunately, I prove instead that $2tdelta(t^2-a^2)=-delta(t+|a|)+delta(t-|a|)$.



          It's easier to approximate $delta$ by $phi_epsilon=epsilon^{-1}phi(t/epsilon)$, where $phi$ is compactly supported and even. Now, let $psi$ be a test function and evaluate
          $$
          begin{align}
          int 2tphi_epsilon(t^2-a^2)psi(t),dt &= int_{-infty}^0+int_0^{infty} \
          &=-int_{-a^2}^infty phi_epsilon(s)psi(-sqrt{s+a^2}),ds +int_{-a^2}^inftyphi_epsilon(s)psi(sqrt{s+a^2}),ds,
          end{align}
          $$
          where I used the change of variables $t^2-a^2=s$. Therefore, taking the limit $epsilonto 0$ we get $2tdelta(t^2-a^2) = -delta(t+|a|)+ delta(t-|a|)$. I don't know if I made a mistake.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 23:45









          user90189user90189

          752616




          752616








          • 1




            I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
            – becko
            Nov 30 '18 at 20:32
















          • 1




            I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
            – becko
            Nov 30 '18 at 20:32










          1




          1




          I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
          – becko
          Nov 30 '18 at 20:32






          I agree, $2lambdadelta(lambda-lambda_n^2) = delta(lambda-lambda_n) + delta(lambda+lambda_n)$. But I think in this case it is safe to assume that $lambda,lambda_n > 0$ (singular values of random matrices are almost surely positive), in which case $delta(lambda+lambda_n) = 0$. But I'm not 100% convinced by this argument.
          – becko
          Nov 30 '18 at 20:32




















          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%2f3019236%2fidentity-between-resolvent-and-singular-value-density%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

          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

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Image
          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
          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