largest singular value of gaussian random matrix
$begingroup$
Let $A = A_{ij}, 1le ile n,1le jle m,$ be a random matrix such that its entries are iid sub-Gaussian random variables with variance proxy $sigma^2$. Show that there exits a constant $C>0$ such that
$$
E||A|| le C(sqrt{m}+sqrt{n}),
$$
where $||A||=sup_{|x|_2=1}|Ax|_2$ is the operator norm of $A$.
This is problem 1.2b in this MIT opencourseware assignment. The result is proven as Theorem 5.32 of these random matrix theory notes, in the special case of gaussian $A_{ij}$, but that result cites another result. Based on the level of the accompanying notes preceding the problem set, I would not expect the cited result to be assumed of the students (I may be wrong of course). So I am wondering about a more direct proof, or whatever the instructor likely had in mind.
matrices probability-theory random-matrices
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
$endgroup$
add a comment |
$begingroup$
Let $A = A_{ij}, 1le ile n,1le jle m,$ be a random matrix such that its entries are iid sub-Gaussian random variables with variance proxy $sigma^2$. Show that there exits a constant $C>0$ such that
$$
E||A|| le C(sqrt{m}+sqrt{n}),
$$
where $||A||=sup_{|x|_2=1}|Ax|_2$ is the operator norm of $A$.
This is problem 1.2b in this MIT opencourseware assignment. The result is proven as Theorem 5.32 of these random matrix theory notes, in the special case of gaussian $A_{ij}$, but that result cites another result. Based on the level of the accompanying notes preceding the problem set, I would not expect the cited result to be assumed of the students (I may be wrong of course). So I am wondering about a more direct proof, or whatever the instructor likely had in mind.
matrices probability-theory random-matrices
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
$endgroup$
add a comment |
$begingroup$
Let $A = A_{ij}, 1le ile n,1le jle m,$ be a random matrix such that its entries are iid sub-Gaussian random variables with variance proxy $sigma^2$. Show that there exits a constant $C>0$ such that
$$
E||A|| le C(sqrt{m}+sqrt{n}),
$$
where $||A||=sup_{|x|_2=1}|Ax|_2$ is the operator norm of $A$.
This is problem 1.2b in this MIT opencourseware assignment. The result is proven as Theorem 5.32 of these random matrix theory notes, in the special case of gaussian $A_{ij}$, but that result cites another result. Based on the level of the accompanying notes preceding the problem set, I would not expect the cited result to be assumed of the students (I may be wrong of course). So I am wondering about a more direct proof, or whatever the instructor likely had in mind.
matrices probability-theory random-matrices
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
$endgroup$
Let $A = A_{ij}, 1le ile n,1le jle m,$ be a random matrix such that its entries are iid sub-Gaussian random variables with variance proxy $sigma^2$. Show that there exits a constant $C>0$ such that
$$
E||A|| le C(sqrt{m}+sqrt{n}),
$$
where $||A||=sup_{|x|_2=1}|Ax|_2$ is the operator norm of $A$.
This is problem 1.2b in this MIT opencourseware assignment. The result is proven as Theorem 5.32 of these random matrix theory notes, in the special case of gaussian $A_{ij}$, but that result cites another result. Based on the level of the accompanying notes preceding the problem set, I would not expect the cited result to be assumed of the students (I may be wrong of course). So I am wondering about a more direct proof, or whatever the instructor likely had in mind.
matrices probability-theory random-matrices
matrices probability-theory random-matrices
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
asked Dec 29 '18 at 21:04
Hasse1987Hasse1987
58148
58148
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
I think you can imitate the proof of Theorem 1.19 from your notes. Apologies if my approach is a little clumsy.
One can show that $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top A v$.
Then $E|A| = E[ sup_{|u|_2le 1, |v|_2 le 1} u^top A v]$.
One can obtain an $1/2$-net $mathcal{N}^n$ over $mathcal{B}_2^n$ with $6^n$ points. Similarly one obtains a $1/2$-net $mathcal{N}^m$ over $mathcal{B}_2^m$ of size $6^m$.
So writing $$u^top A v = (u-x)^top A (v-y) + x^top A v + u^top A y - x^top A y$$ where $x in mathcal{N}^n$, $y in mathcal{N}^m$, and $|x-u|_2 le 1/2$ and $|y-v|_2 le 1/2$ yields
$$E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y]
+ E[sup_{u in mathcal{B}_2^n/2, v in mathcal{B}_2^m/2} u^top A v]. $$
Rearranging leads to
$$frac{3}{4} E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le
E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y].$$
The first term on the right-hand side is the maximum of $6^{n+m}$ sub-Gaussian random variables with variance proxy $sigma^2$, so it is $le sigma sqrt{2 (m+n) log 6}$.
I believe you can bound the other two terms by doing a further net argument and obtaining the same $c sigma sqrt{m+n}$ rate. Finally $sqrt{m+n} le sqrt{m} + sqrt{n}$.
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
add a comment |
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$begingroup$
I think you can imitate the proof of Theorem 1.19 from your notes. Apologies if my approach is a little clumsy.
One can show that $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top A v$.
Then $E|A| = E[ sup_{|u|_2le 1, |v|_2 le 1} u^top A v]$.
One can obtain an $1/2$-net $mathcal{N}^n$ over $mathcal{B}_2^n$ with $6^n$ points. Similarly one obtains a $1/2$-net $mathcal{N}^m$ over $mathcal{B}_2^m$ of size $6^m$.
So writing $$u^top A v = (u-x)^top A (v-y) + x^top A v + u^top A y - x^top A y$$ where $x in mathcal{N}^n$, $y in mathcal{N}^m$, and $|x-u|_2 le 1/2$ and $|y-v|_2 le 1/2$ yields
$$E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y]
+ E[sup_{u in mathcal{B}_2^n/2, v in mathcal{B}_2^m/2} u^top A v]. $$
Rearranging leads to
$$frac{3}{4} E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le
E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y].$$
The first term on the right-hand side is the maximum of $6^{n+m}$ sub-Gaussian random variables with variance proxy $sigma^2$, so it is $le sigma sqrt{2 (m+n) log 6}$.
I believe you can bound the other two terms by doing a further net argument and obtaining the same $c sigma sqrt{m+n}$ rate. Finally $sqrt{m+n} le sqrt{m} + sqrt{n}$.
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
add a comment |
$begingroup$
I think you can imitate the proof of Theorem 1.19 from your notes. Apologies if my approach is a little clumsy.
One can show that $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top A v$.
Then $E|A| = E[ sup_{|u|_2le 1, |v|_2 le 1} u^top A v]$.
One can obtain an $1/2$-net $mathcal{N}^n$ over $mathcal{B}_2^n$ with $6^n$ points. Similarly one obtains a $1/2$-net $mathcal{N}^m$ over $mathcal{B}_2^m$ of size $6^m$.
So writing $$u^top A v = (u-x)^top A (v-y) + x^top A v + u^top A y - x^top A y$$ where $x in mathcal{N}^n$, $y in mathcal{N}^m$, and $|x-u|_2 le 1/2$ and $|y-v|_2 le 1/2$ yields
$$E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y]
+ E[sup_{u in mathcal{B}_2^n/2, v in mathcal{B}_2^m/2} u^top A v]. $$
Rearranging leads to
$$frac{3}{4} E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le
E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y].$$
The first term on the right-hand side is the maximum of $6^{n+m}$ sub-Gaussian random variables with variance proxy $sigma^2$, so it is $le sigma sqrt{2 (m+n) log 6}$.
I believe you can bound the other two terms by doing a further net argument and obtaining the same $c sigma sqrt{m+n}$ rate. Finally $sqrt{m+n} le sqrt{m} + sqrt{n}$.
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
add a comment |
$begingroup$
I think you can imitate the proof of Theorem 1.19 from your notes. Apologies if my approach is a little clumsy.
One can show that $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top A v$.
Then $E|A| = E[ sup_{|u|_2le 1, |v|_2 le 1} u^top A v]$.
One can obtain an $1/2$-net $mathcal{N}^n$ over $mathcal{B}_2^n$ with $6^n$ points. Similarly one obtains a $1/2$-net $mathcal{N}^m$ over $mathcal{B}_2^m$ of size $6^m$.
So writing $$u^top A v = (u-x)^top A (v-y) + x^top A v + u^top A y - x^top A y$$ where $x in mathcal{N}^n$, $y in mathcal{N}^m$, and $|x-u|_2 le 1/2$ and $|y-v|_2 le 1/2$ yields
$$E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y]
+ E[sup_{u in mathcal{B}_2^n/2, v in mathcal{B}_2^m/2} u^top A v]. $$
Rearranging leads to
$$frac{3}{4} E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le
E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y].$$
The first term on the right-hand side is the maximum of $6^{n+m}$ sub-Gaussian random variables with variance proxy $sigma^2$, so it is $le sigma sqrt{2 (m+n) log 6}$.
I believe you can bound the other two terms by doing a further net argument and obtaining the same $c sigma sqrt{m+n}$ rate. Finally $sqrt{m+n} le sqrt{m} + sqrt{n}$.
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
$endgroup$
I think you can imitate the proof of Theorem 1.19 from your notes. Apologies if my approach is a little clumsy.
One can show that $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top A v$.
Then $E|A| = E[ sup_{|u|_2le 1, |v|_2 le 1} u^top A v]$.
One can obtain an $1/2$-net $mathcal{N}^n$ over $mathcal{B}_2^n$ with $6^n$ points. Similarly one obtains a $1/2$-net $mathcal{N}^m$ over $mathcal{B}_2^m$ of size $6^m$.
So writing $$u^top A v = (u-x)^top A (v-y) + x^top A v + u^top A y - x^top A y$$ where $x in mathcal{N}^n$, $y in mathcal{N}^m$, and $|x-u|_2 le 1/2$ and $|y-v|_2 le 1/2$ yields
$$E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y]
+ E[sup_{u in mathcal{B}_2^n/2, v in mathcal{B}_2^m/2} u^top A v]. $$
Rearranging leads to
$$frac{3}{4} E[sup_{u in mathcal{B}_2^n, v in mathcal{B}_2^m} u^top A v]
le
E[sup_{x in mathcal{N}^n, y in mathcal{N}^m} x^top A y]
+ E[sup_{x in mathcal{N}^n, v in mathcal{B}_2^m/2} x^top A v]
+ E[sup_{u in mathcal{B}_2^n/2, y in mathcal{N}^m} u^top A y].$$
The first term on the right-hand side is the maximum of $6^{n+m}$ sub-Gaussian random variables with variance proxy $sigma^2$, so it is $le sigma sqrt{2 (m+n) log 6}$.
I believe you can bound the other two terms by doing a further net argument and obtaining the same $c sigma sqrt{m+n}$ rate. Finally $sqrt{m+n} le sqrt{m} + sqrt{n}$.
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
edited Dec 30 '18 at 0:39
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 22:11
angryavianangryavian
42.2k23481
42.2k23481
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
add a comment |
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
In $|A| = sup_{|u|_2 le 1, |v|_2 le 1} u^top X v$, what is $X$? It should be $A^top A$, shouldn't it? But in your proof, are you taking the quadratic form in $A$?
$endgroup$
– Hasse1987
Dec 29 '18 at 23:12
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
@Hasse1987 Sorry it should be $A$. And no, if it were $A^top A$ that would give you the square of the maximum singular value.
$endgroup$
– angryavian
Dec 30 '18 at 0:40
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
thanks--do you have a reference for that result? The min-max theorem for singular values I see on wikipedia is stated in terms of $A^T A$.
$endgroup$
– Hasse1987
Dec 30 '18 at 0:50
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
$begingroup$
@Hasse1987 See the middle of the first page of this file.
$endgroup$
– angryavian
Dec 30 '18 at 1:16
add a comment |
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