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













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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          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




















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