Are the eigenvectors of real Wigner matrices made of independent random variables with zero-mean?
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I am trying to understand a portion of this paper [p. 3]
and got stuck in the following statement,
which sounded kind of trivial,
but has been deceiving me for a while.
Would you help me understand?
Let
$X$
be an
$n times n$
real symmetric random matrix
with independent entries of zero mean,
$x_1, dots, x_n$
its (unit) eigenvectors,
$A = {1, dots, n/2}$,
$B = {n/2 + 1, dots, n}$
and let
$u$
be the vector where
$u_i = frac{1}{sqrt{n}}$
if $i in A$,
and
$u_i = frac{-1}{sqrt{n}}$
if
$i in B$.
On page 3, left column, the authors claim:
(...) we note that since
$X$
is a random matrix,
its eigenvectors are also random,
so that cross terms cancel
in the quantity
$(u^T x_i)^2$
and the average value is simply
$| x_i |^2/n = 1/n$.
Since
$begin{align}n(u^T x_i)^2 &=
sum_{k=1}^n (x_i)_k^2
+ left( sum_{k neq l; (k, l) in (A times A) cup (B times B)} (x_i)_k (x_i)_l
- sum_{(k, l) in (A times B) cup (B times A)} (x_i)_k (x_i)_l right)\
&= |x_i|^2 + Z_i,end{align}$
we want to show that
$mathbb{E}[Z_i] = 0$.
Now,
it seems the authors are suggesting that since
$X$
is such a matrix,
then when
$n rightarrow infty$
(implied throughout the paper)
we may see the coordinates within each of its eigenvectors as
independent random variables of zero mean.
I can't see how,
since they are so tangled by the elements of the matrix
(as by the spectral theorem we have
$X_{ij} = sum_{k=1}^n lambda_k (x_i)_k (x_j)_k$).
Maybe the answer is not as general as it seems
and relies on features of the specific distribution of this
$X$.
The
$X_{ij}$
are independent Bernoulli-distributed variables centered at the mean, with
$X_{ij} sim
begin{cases}
Be(p) - p text{, if } {i,j} in (A times A) cup (B times B) \
Be(q) - q text{, if } {i,j} in (A times B) cup (B times A).
end{cases}$
The argument they refer to [p. 507] solves a similar case,
where they were considering a vector
$v$
drawn uniformly from the unit sphere
and were concerned with
$|v|^2$
instead of
$(u^T x_i)^2$.
After some research
(e.g. this survey [§2 of Ch. 6, on p. 15]
on eigenvectors of random matrices),
it seems that for Wigner matrices
like the one I am considering,
it is expected that the eigenvectors follow
a distribution close to uniform from the unit sphere
(a... Haar measure?),
so maybe this is the missing link?
But then how do
Bernoulli-distributed variables
centered at the mean
relate to,
for example,
the Gaussian Orthogonal Ensemble
or anything like that?
(I could not find anything,
since the first four moments don't match
the criteria for direct comparisons).
I have been trying to see if
there is something about the adjacency matrix
of random graphs on this matter as well,
but it has been unfruitful up to now.
Am I on the right track, even?
If you read this up to here, thank you for reading! (:
linear-algebra matrices symmetric-matrices random-graphs random-matrices
add a comment |
up vote
2
down vote
favorite
I am trying to understand a portion of this paper [p. 3]
and got stuck in the following statement,
which sounded kind of trivial,
but has been deceiving me for a while.
Would you help me understand?
Let
$X$
be an
$n times n$
real symmetric random matrix
with independent entries of zero mean,
$x_1, dots, x_n$
its (unit) eigenvectors,
$A = {1, dots, n/2}$,
$B = {n/2 + 1, dots, n}$
and let
$u$
be the vector where
$u_i = frac{1}{sqrt{n}}$
if $i in A$,
and
$u_i = frac{-1}{sqrt{n}}$
if
$i in B$.
On page 3, left column, the authors claim:
(...) we note that since
$X$
is a random matrix,
its eigenvectors are also random,
so that cross terms cancel
in the quantity
$(u^T x_i)^2$
and the average value is simply
$| x_i |^2/n = 1/n$.
Since
$begin{align}n(u^T x_i)^2 &=
sum_{k=1}^n (x_i)_k^2
+ left( sum_{k neq l; (k, l) in (A times A) cup (B times B)} (x_i)_k (x_i)_l
- sum_{(k, l) in (A times B) cup (B times A)} (x_i)_k (x_i)_l right)\
&= |x_i|^2 + Z_i,end{align}$
we want to show that
$mathbb{E}[Z_i] = 0$.
Now,
it seems the authors are suggesting that since
$X$
is such a matrix,
then when
$n rightarrow infty$
(implied throughout the paper)
we may see the coordinates within each of its eigenvectors as
independent random variables of zero mean.
I can't see how,
since they are so tangled by the elements of the matrix
(as by the spectral theorem we have
$X_{ij} = sum_{k=1}^n lambda_k (x_i)_k (x_j)_k$).
Maybe the answer is not as general as it seems
and relies on features of the specific distribution of this
$X$.
The
$X_{ij}$
are independent Bernoulli-distributed variables centered at the mean, with
$X_{ij} sim
begin{cases}
Be(p) - p text{, if } {i,j} in (A times A) cup (B times B) \
Be(q) - q text{, if } {i,j} in (A times B) cup (B times A).
end{cases}$
The argument they refer to [p. 507] solves a similar case,
where they were considering a vector
$v$
drawn uniformly from the unit sphere
and were concerned with
$|v|^2$
instead of
$(u^T x_i)^2$.
After some research
(e.g. this survey [§2 of Ch. 6, on p. 15]
on eigenvectors of random matrices),
it seems that for Wigner matrices
like the one I am considering,
it is expected that the eigenvectors follow
a distribution close to uniform from the unit sphere
(a... Haar measure?),
so maybe this is the missing link?
But then how do
Bernoulli-distributed variables
centered at the mean
relate to,
for example,
the Gaussian Orthogonal Ensemble
or anything like that?
(I could not find anything,
since the first four moments don't match
the criteria for direct comparisons).
I have been trying to see if
there is something about the adjacency matrix
of random graphs on this matter as well,
but it has been unfruitful up to now.
Am I on the right track, even?
If you read this up to here, thank you for reading! (:
linear-algebra matrices symmetric-matrices random-graphs random-matrices
2
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to understand a portion of this paper [p. 3]
and got stuck in the following statement,
which sounded kind of trivial,
but has been deceiving me for a while.
Would you help me understand?
Let
$X$
be an
$n times n$
real symmetric random matrix
with independent entries of zero mean,
$x_1, dots, x_n$
its (unit) eigenvectors,
$A = {1, dots, n/2}$,
$B = {n/2 + 1, dots, n}$
and let
$u$
be the vector where
$u_i = frac{1}{sqrt{n}}$
if $i in A$,
and
$u_i = frac{-1}{sqrt{n}}$
if
$i in B$.
On page 3, left column, the authors claim:
(...) we note that since
$X$
is a random matrix,
its eigenvectors are also random,
so that cross terms cancel
in the quantity
$(u^T x_i)^2$
and the average value is simply
$| x_i |^2/n = 1/n$.
Since
$begin{align}n(u^T x_i)^2 &=
sum_{k=1}^n (x_i)_k^2
+ left( sum_{k neq l; (k, l) in (A times A) cup (B times B)} (x_i)_k (x_i)_l
- sum_{(k, l) in (A times B) cup (B times A)} (x_i)_k (x_i)_l right)\
&= |x_i|^2 + Z_i,end{align}$
we want to show that
$mathbb{E}[Z_i] = 0$.
Now,
it seems the authors are suggesting that since
$X$
is such a matrix,
then when
$n rightarrow infty$
(implied throughout the paper)
we may see the coordinates within each of its eigenvectors as
independent random variables of zero mean.
I can't see how,
since they are so tangled by the elements of the matrix
(as by the spectral theorem we have
$X_{ij} = sum_{k=1}^n lambda_k (x_i)_k (x_j)_k$).
Maybe the answer is not as general as it seems
and relies on features of the specific distribution of this
$X$.
The
$X_{ij}$
are independent Bernoulli-distributed variables centered at the mean, with
$X_{ij} sim
begin{cases}
Be(p) - p text{, if } {i,j} in (A times A) cup (B times B) \
Be(q) - q text{, if } {i,j} in (A times B) cup (B times A).
end{cases}$
The argument they refer to [p. 507] solves a similar case,
where they were considering a vector
$v$
drawn uniformly from the unit sphere
and were concerned with
$|v|^2$
instead of
$(u^T x_i)^2$.
After some research
(e.g. this survey [§2 of Ch. 6, on p. 15]
on eigenvectors of random matrices),
it seems that for Wigner matrices
like the one I am considering,
it is expected that the eigenvectors follow
a distribution close to uniform from the unit sphere
(a... Haar measure?),
so maybe this is the missing link?
But then how do
Bernoulli-distributed variables
centered at the mean
relate to,
for example,
the Gaussian Orthogonal Ensemble
or anything like that?
(I could not find anything,
since the first four moments don't match
the criteria for direct comparisons).
I have been trying to see if
there is something about the adjacency matrix
of random graphs on this matter as well,
but it has been unfruitful up to now.
Am I on the right track, even?
If you read this up to here, thank you for reading! (:
linear-algebra matrices symmetric-matrices random-graphs random-matrices
I am trying to understand a portion of this paper [p. 3]
and got stuck in the following statement,
which sounded kind of trivial,
but has been deceiving me for a while.
Would you help me understand?
Let
$X$
be an
$n times n$
real symmetric random matrix
with independent entries of zero mean,
$x_1, dots, x_n$
its (unit) eigenvectors,
$A = {1, dots, n/2}$,
$B = {n/2 + 1, dots, n}$
and let
$u$
be the vector where
$u_i = frac{1}{sqrt{n}}$
if $i in A$,
and
$u_i = frac{-1}{sqrt{n}}$
if
$i in B$.
On page 3, left column, the authors claim:
(...) we note that since
$X$
is a random matrix,
its eigenvectors are also random,
so that cross terms cancel
in the quantity
$(u^T x_i)^2$
and the average value is simply
$| x_i |^2/n = 1/n$.
Since
$begin{align}n(u^T x_i)^2 &=
sum_{k=1}^n (x_i)_k^2
+ left( sum_{k neq l; (k, l) in (A times A) cup (B times B)} (x_i)_k (x_i)_l
- sum_{(k, l) in (A times B) cup (B times A)} (x_i)_k (x_i)_l right)\
&= |x_i|^2 + Z_i,end{align}$
we want to show that
$mathbb{E}[Z_i] = 0$.
Now,
it seems the authors are suggesting that since
$X$
is such a matrix,
then when
$n rightarrow infty$
(implied throughout the paper)
we may see the coordinates within each of its eigenvectors as
independent random variables of zero mean.
I can't see how,
since they are so tangled by the elements of the matrix
(as by the spectral theorem we have
$X_{ij} = sum_{k=1}^n lambda_k (x_i)_k (x_j)_k$).
Maybe the answer is not as general as it seems
and relies on features of the specific distribution of this
$X$.
The
$X_{ij}$
are independent Bernoulli-distributed variables centered at the mean, with
$X_{ij} sim
begin{cases}
Be(p) - p text{, if } {i,j} in (A times A) cup (B times B) \
Be(q) - q text{, if } {i,j} in (A times B) cup (B times A).
end{cases}$
The argument they refer to [p. 507] solves a similar case,
where they were considering a vector
$v$
drawn uniformly from the unit sphere
and were concerned with
$|v|^2$
instead of
$(u^T x_i)^2$.
After some research
(e.g. this survey [§2 of Ch. 6, on p. 15]
on eigenvectors of random matrices),
it seems that for Wigner matrices
like the one I am considering,
it is expected that the eigenvectors follow
a distribution close to uniform from the unit sphere
(a... Haar measure?),
so maybe this is the missing link?
But then how do
Bernoulli-distributed variables
centered at the mean
relate to,
for example,
the Gaussian Orthogonal Ensemble
or anything like that?
(I could not find anything,
since the first four moments don't match
the criteria for direct comparisons).
I have been trying to see if
there is something about the adjacency matrix
of random graphs on this matter as well,
but it has been unfruitful up to now.
Am I on the right track, even?
If you read this up to here, thank you for reading! (:
linear-algebra matrices symmetric-matrices random-graphs random-matrices
linear-algebra matrices symmetric-matrices random-graphs random-matrices
edited Nov 26 at 17:32
asked Nov 24 at 1:08
Felix Liu
214
214
2
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40
add a comment |
2
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40
2
2
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40
add a comment |
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2
Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif{3,-1,-2}$, we have $E((u^Tx_1)^2)=0.60846nefrac12$.
– user1551
Nov 26 at 20:15
@user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n to infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet...
– Felix Liu
Nov 27 at 2:40