Relation between Frobenius, spectral norm and sum of maxima












4












$begingroup$


Let $A$ be a $n times n$ matrix so that the Frobenius norm squared $|A|_F^2$ is $Theta(n)$, the spectral norm squared $|A|_2^2=1$. Is it true that $sum_{i=1}^nmax_{1leq jleq n} |A_{ij}|^2$ is $Omega(n)$? Assume that $n$ is sufficiently large.



I cannot find a relation between matrix norms that can show this. The idea behind this question is that there are many singular values of $A$ that are $Theta(1)$.



Thanks!






linear-algebra matrices norms







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  • 2




    $begingroup$
    According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
    $endgroup$
    – W-t-P
    yesterday






  • 1




    $begingroup$
    Frobenius norm, where did you find that? It is wrong what you are saying.
    $endgroup$
    – horxio
    yesterday








  • 1




    $begingroup$
    What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
    $endgroup$
    – horxio
    yesterday






  • 2




    $begingroup$
    The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
    $endgroup$
    – W-t-P
    yesterday








  • 2




    $begingroup$
    @W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
    $endgroup$
    – Federico Poloni
    yesterday
















4












$begingroup$


Let $A$ be a $n times n$ matrix so that the Frobenius norm squared $|A|_F^2$ is $Theta(n)$, the spectral norm squared $|A|_2^2=1$. Is it true that $sum_{i=1}^nmax_{1leq jleq n} |A_{ij}|^2$ is $Omega(n)$? Assume that $n$ is sufficiently large.



I cannot find a relation between matrix norms that can show this. The idea behind this question is that there are many singular values of $A$ that are $Theta(1)$.



Thanks!






linear-algebra matrices norms







share|cite|improve this question









New contributor




horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
    $endgroup$
    – W-t-P
    yesterday






  • 1




    $begingroup$
    Frobenius norm, where did you find that? It is wrong what you are saying.
    $endgroup$
    – horxio
    yesterday








  • 1




    $begingroup$
    What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
    $endgroup$
    – horxio
    yesterday






  • 2




    $begingroup$
    The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
    $endgroup$
    – W-t-P
    yesterday








  • 2




    $begingroup$
    @W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
    $endgroup$
    – Federico Poloni
    yesterday














4












4








4





$begingroup$


Let $A$ be a $n times n$ matrix so that the Frobenius norm squared $|A|_F^2$ is $Theta(n)$, the spectral norm squared $|A|_2^2=1$. Is it true that $sum_{i=1}^nmax_{1leq jleq n} |A_{ij}|^2$ is $Omega(n)$? Assume that $n$ is sufficiently large.



I cannot find a relation between matrix norms that can show this. The idea behind this question is that there are many singular values of $A$ that are $Theta(1)$.



Thanks!






linear-algebra matrices norms







share|cite|improve this question









New contributor




horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $A$ be a $n times n$ matrix so that the Frobenius norm squared $|A|_F^2$ is $Theta(n)$, the spectral norm squared $|A|_2^2=1$. Is it true that $sum_{i=1}^nmax_{1leq jleq n} |A_{ij}|^2$ is $Omega(n)$? Assume that $n$ is sufficiently large.



I cannot find a relation between matrix norms that can show this. The idea behind this question is that there are many singular values of $A$ that are $Theta(1)$.



Thanks!






linear-algebra matrices norms




linear-algebra matrices norms






share|cite|improve this question









New contributor




horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Liviu Nicolaescu

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asked yesterday









horxiohorxio

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New contributor




horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






horxio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    $begingroup$
    According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
    $endgroup$
    – W-t-P
    yesterday






  • 1




    $begingroup$
    Frobenius norm, where did you find that? It is wrong what you are saying.
    $endgroup$
    – horxio
    yesterday








  • 1




    $begingroup$
    What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
    $endgroup$
    – horxio
    yesterday






  • 2




    $begingroup$
    The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
    $endgroup$
    – W-t-P
    yesterday








  • 2




    $begingroup$
    @W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
    $endgroup$
    – Federico Poloni
    yesterday














  • 2




    $begingroup$
    According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
    $endgroup$
    – W-t-P
    yesterday






  • 1




    $begingroup$
    Frobenius norm, where did you find that? It is wrong what you are saying.
    $endgroup$
    – horxio
    yesterday








  • 1




    $begingroup$
    What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
    $endgroup$
    – horxio
    yesterday






  • 2




    $begingroup$
    The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
    $endgroup$
    – W-t-P
    yesterday








  • 2




    $begingroup$
    @W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
    $endgroup$
    – Federico Poloni
    yesterday








2




2




$begingroup$
According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
$endgroup$
– W-t-P
yesterday




$begingroup$
According to Wikipedia, $|A|_F=|A|_2$. Please, explain your notation!
$endgroup$
– W-t-P
yesterday




1




1




$begingroup$
Frobenius norm, where did you find that? It is wrong what you are saying.
$endgroup$
– horxio
yesterday






$begingroup$
Frobenius norm, where did you find that? It is wrong what you are saying.
$endgroup$
– horxio
yesterday






1




1




$begingroup$
What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
$endgroup$
– horxio
yesterday




$begingroup$
What you are saying is incorrect. Can you please tell me where exactly you found this relation? It is wrong that the Frobenius norm is equal to the spectral norm. Think about it, if they were equal, why should we have two definitions? it holds that $||A||_F geq ||A||_2$.
$endgroup$
– horxio
yesterday




2




2




$begingroup$
The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
$endgroup$
– W-t-P
yesterday






$begingroup$
The last sentence in the ""Entrywise" matrix norms" reads: The special case p = 2 is the Frobenius norm; see also the "Frobenius norm" section below. Instead of arguing who is (in)correct, please explain your notation.
$endgroup$
– W-t-P
yesterday






2




2




$begingroup$
@W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
$endgroup$
– Federico Poloni
yesterday




$begingroup$
@W-t-P I find your comments towards a new user a bit aggressive. The problem here is that $|A|_2$ is standard notation for two different things, as the Wikipedia page that you linked also notes (if you read a bit earlier, These norms again share the notation with the induced and entrywise p-norms, but they are different, and earlier the definition of the spectral norm). On the other hand, the terms Frobenius norm and spectral norm are unambiguous and look perfectly fine to me as explanations of the notation in OP's question.
$endgroup$
– Federico Poloni
yesterday










1 Answer
1






active

oldest

votes


















0












$begingroup$

This is false in general, but true for matrices with non-negative entries.



For a counterexample, suppose that $n=p$ is prime, and consider the matrix
$$ A=left|p^{-1/2}left(frac{i-j}pright)right|_{i,j=0,dotsc,p-1} $$
where $(cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normzlized Gaussian sums, equal $1$ in absolute value; hence, $|A|_2le 1$. Also, we have $|A|_F^2=p-1$. On the other hand,
$$ sum_i max_j |A_{ij}|^2 = 1. $$



Suppose now that all elements of $A$ are non-negative. Let $u_iin{mathbb R}^n$ be the row vectors of $A$, and denote by $|cdot|_p$ the $ell^p$-norm over ${mathbb R}^n$; when $p=2$, this is the standard Euclidian norm. The Frobenius norm of $A$ is $|A|_F^2=sum_i|u_i|_2^2$. Assuming that $|A|_F^2ge cn$ and $|A|_2^2le C$, we show that $sum_i|u_i|_infty^2ge C^{-1}c^2n$.



Denoting by $vec 1$ the all-$1$ vector, we have
$$ C ge |A|_2^2 = max_x frac{|Ax|_2^2}{|x|_2^2} ge frac{|Avec 1|_2^2}{|vec 1|_2^2} = frac1nsum_i |u_i|_1^2. $$
(It is this computation that uses the non-negativeness assumption.) This
implies
$$ sum_i |u_i|_1^2 le Cn $$
and, consequently, by Cauchy-Schwarz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
    $endgroup$
    – horxio
    8 hours ago










  • $begingroup$
    @horxio: Hope everything is corrected now.
    $endgroup$
    – Seva
    2 hours ago










  • $begingroup$
    Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
    $endgroup$
    – horxio
    34 mins ago












  • $begingroup$
    @horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
    $endgroup$
    – Seva
    8 mins ago












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

This is false in general, but true for matrices with non-negative entries.



For a counterexample, suppose that $n=p$ is prime, and consider the matrix
$$ A=left|p^{-1/2}left(frac{i-j}pright)right|_{i,j=0,dotsc,p-1} $$
where $(cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normzlized Gaussian sums, equal $1$ in absolute value; hence, $|A|_2le 1$. Also, we have $|A|_F^2=p-1$. On the other hand,
$$ sum_i max_j |A_{ij}|^2 = 1. $$



Suppose now that all elements of $A$ are non-negative. Let $u_iin{mathbb R}^n$ be the row vectors of $A$, and denote by $|cdot|_p$ the $ell^p$-norm over ${mathbb R}^n$; when $p=2$, this is the standard Euclidian norm. The Frobenius norm of $A$ is $|A|_F^2=sum_i|u_i|_2^2$. Assuming that $|A|_F^2ge cn$ and $|A|_2^2le C$, we show that $sum_i|u_i|_infty^2ge C^{-1}c^2n$.



Denoting by $vec 1$ the all-$1$ vector, we have
$$ C ge |A|_2^2 = max_x frac{|Ax|_2^2}{|x|_2^2} ge frac{|Avec 1|_2^2}{|vec 1|_2^2} = frac1nsum_i |u_i|_1^2. $$
(It is this computation that uses the non-negativeness assumption.) This
implies
$$ sum_i |u_i|_1^2 le Cn $$
and, consequently, by Cauchy-Schwarz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
    $endgroup$
    – horxio
    8 hours ago










  • $begingroup$
    @horxio: Hope everything is corrected now.
    $endgroup$
    – Seva
    2 hours ago










  • $begingroup$
    Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
    $endgroup$
    – horxio
    34 mins ago












  • $begingroup$
    @horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
    $endgroup$
    – Seva
    8 mins ago
















0












$begingroup$

This is false in general, but true for matrices with non-negative entries.



For a counterexample, suppose that $n=p$ is prime, and consider the matrix
$$ A=left|p^{-1/2}left(frac{i-j}pright)right|_{i,j=0,dotsc,p-1} $$
where $(cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normzlized Gaussian sums, equal $1$ in absolute value; hence, $|A|_2le 1$. Also, we have $|A|_F^2=p-1$. On the other hand,
$$ sum_i max_j |A_{ij}|^2 = 1. $$



Suppose now that all elements of $A$ are non-negative. Let $u_iin{mathbb R}^n$ be the row vectors of $A$, and denote by $|cdot|_p$ the $ell^p$-norm over ${mathbb R}^n$; when $p=2$, this is the standard Euclidian norm. The Frobenius norm of $A$ is $|A|_F^2=sum_i|u_i|_2^2$. Assuming that $|A|_F^2ge cn$ and $|A|_2^2le C$, we show that $sum_i|u_i|_infty^2ge C^{-1}c^2n$.



Denoting by $vec 1$ the all-$1$ vector, we have
$$ C ge |A|_2^2 = max_x frac{|Ax|_2^2}{|x|_2^2} ge frac{|Avec 1|_2^2}{|vec 1|_2^2} = frac1nsum_i |u_i|_1^2. $$
(It is this computation that uses the non-negativeness assumption.) This
implies
$$ sum_i |u_i|_1^2 le Cn $$
and, consequently, by Cauchy-Schwarz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
    $endgroup$
    – horxio
    8 hours ago










  • $begingroup$
    @horxio: Hope everything is corrected now.
    $endgroup$
    – Seva
    2 hours ago










  • $begingroup$
    Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
    $endgroup$
    – horxio
    34 mins ago












  • $begingroup$
    @horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
    $endgroup$
    – Seva
    8 mins ago














0












0








0





$begingroup$

This is false in general, but true for matrices with non-negative entries.



For a counterexample, suppose that $n=p$ is prime, and consider the matrix
$$ A=left|p^{-1/2}left(frac{i-j}pright)right|_{i,j=0,dotsc,p-1} $$
where $(cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normzlized Gaussian sums, equal $1$ in absolute value; hence, $|A|_2le 1$. Also, we have $|A|_F^2=p-1$. On the other hand,
$$ sum_i max_j |A_{ij}|^2 = 1. $$



Suppose now that all elements of $A$ are non-negative. Let $u_iin{mathbb R}^n$ be the row vectors of $A$, and denote by $|cdot|_p$ the $ell^p$-norm over ${mathbb R}^n$; when $p=2$, this is the standard Euclidian norm. The Frobenius norm of $A$ is $|A|_F^2=sum_i|u_i|_2^2$. Assuming that $|A|_F^2ge cn$ and $|A|_2^2le C$, we show that $sum_i|u_i|_infty^2ge C^{-1}c^2n$.



Denoting by $vec 1$ the all-$1$ vector, we have
$$ C ge |A|_2^2 = max_x frac{|Ax|_2^2}{|x|_2^2} ge frac{|Avec 1|_2^2}{|vec 1|_2^2} = frac1nsum_i |u_i|_1^2. $$
(It is this computation that uses the non-negativeness assumption.) This
implies
$$ sum_i |u_i|_1^2 le Cn $$
and, consequently, by Cauchy-Schwarz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$






share|cite|improve this answer











$endgroup$



This is false in general, but true for matrices with non-negative entries.



For a counterexample, suppose that $n=p$ is prime, and consider the matrix
$$ A=left|p^{-1/2}left(frac{i-j}pright)right|_{i,j=0,dotsc,p-1} $$
where $(cdot/p)$ is the Legendre symbol. This is a circulant matrix; its non-zero eigenvalues are normzlized Gaussian sums, equal $1$ in absolute value; hence, $|A|_2le 1$. Also, we have $|A|_F^2=p-1$. On the other hand,
$$ sum_i max_j |A_{ij}|^2 = 1. $$



Suppose now that all elements of $A$ are non-negative. Let $u_iin{mathbb R}^n$ be the row vectors of $A$, and denote by $|cdot|_p$ the $ell^p$-norm over ${mathbb R}^n$; when $p=2$, this is the standard Euclidian norm. The Frobenius norm of $A$ is $|A|_F^2=sum_i|u_i|_2^2$. Assuming that $|A|_F^2ge cn$ and $|A|_2^2le C$, we show that $sum_i|u_i|_infty^2ge C^{-1}c^2n$.



Denoting by $vec 1$ the all-$1$ vector, we have
$$ C ge |A|_2^2 = max_x frac{|Ax|_2^2}{|x|_2^2} ge frac{|Avec 1|_2^2}{|vec 1|_2^2} = frac1nsum_i |u_i|_1^2. $$
(It is this computation that uses the non-negativeness assumption.) This
implies
$$ sum_i |u_i|_1^2 le Cn $$
and, consequently, by Cauchy-Schwarz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 9 hours ago









SevaSeva

12.8k138103




12.8k138103












  • $begingroup$
    Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
    $endgroup$
    – horxio
    8 hours ago










  • $begingroup$
    @horxio: Hope everything is corrected now.
    $endgroup$
    – Seva
    2 hours ago










  • $begingroup$
    Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
    $endgroup$
    – horxio
    34 mins ago












  • $begingroup$
    @horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
    $endgroup$
    – Seva
    8 mins ago


















  • $begingroup$
    Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
    $endgroup$
    – horxio
    8 hours ago










  • $begingroup$
    @horxio: Hope everything is corrected now.
    $endgroup$
    – Seva
    2 hours ago










  • $begingroup$
    Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
    $endgroup$
    – horxio
    34 mins ago












  • $begingroup$
    @horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
    $endgroup$
    – Seva
    8 mins ago
















$begingroup$
Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
$endgroup$
– horxio
8 hours ago




$begingroup$
Hi Seva, can you explain the equality $||A vec{1}||_2^2 = sum_i ||u_i||_1^2$. It feels that you have sum of entries of $A$ v.s sum of absolute values of entries of $A$.
$endgroup$
– horxio
8 hours ago












$begingroup$
@horxio: Hope everything is corrected now.
$endgroup$
– Seva
2 hours ago




$begingroup$
@horxio: Hope everything is corrected now.
$endgroup$
– Seva
2 hours ago












$begingroup$
Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
$endgroup$
– horxio
34 mins ago






$begingroup$
Hi Seva, I am not sure about your argument that the spectral norm is bounded by 1. I understand the construction (I guess any circulant matrix with half $frac{1}{sqrt{n}}$ and $-frac{1}{sqrt{n}}$ would work if your argument is correct) but why the eigenvalues are bounded by 1? Say that $p=4k+1$ so that your proposed matrix is symmetric. What is the argument that each eigenvalue is bounded by 1?
$endgroup$
– horxio
34 mins ago














$begingroup$
@horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
$endgroup$
– Seva
8 mins ago




$begingroup$
@horxio: In fact, there is one zero eigenvalue, and the rest are equal to either $pm 1$, or $pm i$, depending on whether $pequiv 1pmod 4$ or $pequiv 3pmod 4$. Check en.wikipedia.org/wiki/… and use the fact that the sums emerging are Gaussian sums. (You can also use Maple / Mathematica / ... to verify this numerically for small values of $p$.)
$endgroup$
– Seva
8 mins ago










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