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
    2 days ago






  • 1




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








  • 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
    2 days ago






  • 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
    2 days ago








  • 3




    $begingroup$
    @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
    2 days ago
















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
    2 days ago






  • 1




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








  • 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
    2 days ago






  • 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
    2 days ago








  • 3




    $begingroup$
    @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
    2 days ago














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 2 days ago









Liviu Nicolaescu

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asked 2 days ago









horxiohorxio

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333




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
    2 days ago






  • 1




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








  • 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
    2 days ago






  • 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
    2 days ago








  • 3




    $begingroup$
    @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
    2 days ago














  • 2




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






  • 1




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








  • 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
    2 days ago






  • 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
    2 days ago








  • 3




    $begingroup$
    @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
    2 days ago








2




2




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




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




1




1




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






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






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
2 days ago




$begingroup$
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
2 days ago




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
2 days ago






$begingroup$
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
2 days ago






3




3




$begingroup$
@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
2 days ago




$begingroup$
@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
2 days ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 normalized 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 Euclidean 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-Schwartz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
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, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    I meant $||A||_2^2$ is $Omega(sqrt{n})$.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    (This answer has been completely changed, it should be correct now.)
    $endgroup$
    – Seva
    13 hours ago












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

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 normalized 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 Euclidean 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-Schwartz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
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, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    I meant $||A||_2^2$ is $Omega(sqrt{n})$.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    (This answer has been completely changed, it should be correct now.)
    $endgroup$
    – Seva
    13 hours ago
















4












$begingroup$

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 normalized 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 Euclidean 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-Schwartz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
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, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    I meant $||A||_2^2$ is $Omega(sqrt{n})$.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    (This answer has been completely changed, it should be correct now.)
    $endgroup$
    – Seva
    13 hours ago














4












4








4





$begingroup$

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 normalized 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 Euclidean 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-Schwartz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
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 normalized 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 Euclidean 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-Schwartz,
$$ cn le |A|_F^2 = sum_i |u_i|_2^2 le sum_i |u_i|_infty |u_i|_1 le left( sum_i |u_i|_infty^2right)^{1/2} left( sum_i |u_i|_1^2right)^{1/2} le
left( Cnsum_i |u_i|_infty^2right)^{1/2}, $$
which yields the desired estimate
$$ sum_i |u_i|_infty^2 ge C^{-1}c^2n. $$







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edited 13 hours ago

























answered 2 days ago









SevaSeva

12.8k138103




12.8k138103












  • $begingroup$
    Hi Seva, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    I meant $||A||_2^2$ is $Omega(sqrt{n})$.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    (This answer has been completely changed, it should be correct now.)
    $endgroup$
    – Seva
    13 hours ago


















  • $begingroup$
    Hi Seva, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    I meant $||A||_2^2$ is $Omega(sqrt{n})$.
    $endgroup$
    – horxio
    2 days ago












  • $begingroup$
    (This answer has been completely changed, it should be correct now.)
    $endgroup$
    – Seva
    13 hours ago
















$begingroup$
Hi Seva, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
$endgroup$
– horxio
2 days ago






$begingroup$
Hi Seva, basically the example you provided is wrong since $||A||_2 geq sqrt{n}gg 1$. Your example is about the spectral radius but not the spectral norm.
$endgroup$
– horxio
2 days ago














$begingroup$
I meant $||A||_2^2$ is $Omega(sqrt{n})$.
$endgroup$
– horxio
2 days ago






$begingroup$
I meant $||A||_2^2$ is $Omega(sqrt{n})$.
$endgroup$
– horxio
2 days ago














$begingroup$
(This answer has been completely changed, it should be correct now.)
$endgroup$
– Seva
13 hours ago




$begingroup$
(This answer has been completely changed, it should be correct now.)
$endgroup$
– Seva
13 hours ago










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