Identity between resolvent and singular value density
$begingroup$
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):
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:
So how does (3) follow from (2)?
linear-algebra random-matrices singularvalues
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
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):
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:
So how does (3) follow from (2)?
linear-algebra random-matrices singularvalues
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
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):
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:
So how does (3) follow from (2)?
linear-algebra random-matrices singularvalues
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
$endgroup$
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):
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:
So how does (3) follow from (2)?
linear-algebra random-matrices singularvalues
linear-algebra random-matrices singularvalues
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
asked Nov 29 '18 at 21:19
beckobecko
2,34931942
2,34931942
add a comment |
add a comment |
1 Answer
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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.
answered Nov 29 '18 at 23:45
user90189user90189
752616
$endgroup$
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
add a comment |
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$begingroup$
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.
answered Nov 29 '18 at 23:45
user90189user90189
752616
$endgroup$
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
add a comment |
$begingroup$
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.
answered Nov 29 '18 at 23:45
user90189user90189
752616
$endgroup$
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
add a comment |
$begingroup$
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.
answered Nov 29 '18 at 23:45
user90189user90189
752616
$endgroup$
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.
answered Nov 29 '18 at 23:45
user90189user90189
752616
answered Nov 29 '18 at 23:45
user90189user90189
752616
answered Nov 29 '18 at 23:45
user90189user90189
752616
answered Nov 29 '18 at 23:45
user90189user90189
752616
752616
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
add a comment |
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
1
1
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
$begingroup$
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.
$endgroup$
– becko
Nov 30 '18 at 20:32
add a comment |
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