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Lower bound spectral radius of matrix
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Let $alpha, beta, gamma in (0, 1)$ such that $alpha < beta, gamma < beta$. Let $d in mathbbN, d geq 4$.
I am interested in bounding from below the spectral radius $rho(K)$ of the following matrix $K$ of size $d$ (or even better determining it analytically).
$$K = beginpmatrix
sqrt(1 - alpha)(1 - gamma) & fracsqrtalpha gammad-1 & cdots & cdots & fracsqrtalpha gammad-1 \
fracsqrtalpha gammad-1 & eta & fracbetad-2 & cdots & fracbetad-2 \
fracsqrtalpha gammad-1 & fracbetad-2 & eta & ddots & vdots \
vdots & vdots & ddots & ddots & fracbetad-2\
fracsqrtalpha gammad-1 & fracbetad-2 & cdots & fracbetad-2 & eta \
endpmatrix$$
where $eta = sqrtleft(1 - beta - fracalphad-1right)left((1 - beta - fracgammad-1right)$
Notice, as it can only help, that $K$ is symmetric, and has strictly positive entries. So far, have only been able to say that the spectral radius lies (up to reordering of the bounds) in
$$left[ sqrt(1 - alpha)(1 - gamma) + sqrtalphagamma; fracsqrtalpha gammad-1+ eta + beta right]$$
since by positivity $rho(K)$ stands in between the minimum and maximum row sums.
Numerically, though, it seems that I have the following dependency $rho(K) = 1 - Oleft( fracf(alpha, gamma)d right)$, and this sort of dependency in $d$ is my goal, but I am a little bit short on tools to confirm it.
linear-algebra eigenvalues-eigenvectors symmetric-matrices upper-lower-bounds spectral-radius
asked Aug 2 at 13:01
ippiki-ookami
303116
add a comment |Â
up vote
0
down vote
favorite
Let $alpha, beta, gamma in (0, 1)$ such that $alpha < beta, gamma < beta$. Let $d in mathbbN, d geq 4$.
I am interested in bounding from below the spectral radius $rho(K)$ of the following matrix $K$ of size $d$ (or even better determining it analytically).
$$K = beginpmatrix
sqrt(1 - alpha)(1 - gamma) & fracsqrtalpha gammad-1 & cdots & cdots & fracsqrtalpha gammad-1 \
fracsqrtalpha gammad-1 & eta & fracbetad-2 & cdots & fracbetad-2 \
fracsqrtalpha gammad-1 & fracbetad-2 & eta & ddots & vdots \
vdots & vdots & ddots & ddots & fracbetad-2\
fracsqrtalpha gammad-1 & fracbetad-2 & cdots & fracbetad-2 & eta \
endpmatrix$$
where $eta = sqrtleft(1 - beta - fracalphad-1right)left((1 - beta - fracgammad-1right)$
Notice, as it can only help, that $K$ is symmetric, and has strictly positive entries. So far, have only been able to say that the spectral radius lies (up to reordering of the bounds) in
$$left[ sqrt(1 - alpha)(1 - gamma) + sqrtalphagamma; fracsqrtalpha gammad-1+ eta + beta right]$$
since by positivity $rho(K)$ stands in between the minimum and maximum row sums.
Numerically, though, it seems that I have the following dependency $rho(K) = 1 - Oleft( fracf(alpha, gamma)d right)$, and this sort of dependency in $d$ is my goal, but I am a little bit short on tools to confirm it.
linear-algebra eigenvalues-eigenvectors symmetric-matrices upper-lower-bounds spectral-radius
asked Aug 2 at 13:01
ippiki-ookami
303116
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $alpha, beta, gamma in (0, 1)$ such that $alpha < beta, gamma < beta$. Let $d in mathbbN, d geq 4$.
I am interested in bounding from below the spectral radius $rho(K)$ of the following matrix $K$ of size $d$ (or even better determining it analytically).
$$K = beginpmatrix
sqrt(1 - alpha)(1 - gamma) & fracsqrtalpha gammad-1 & cdots & cdots & fracsqrtalpha gammad-1 \
fracsqrtalpha gammad-1 & eta & fracbetad-2 & cdots & fracbetad-2 \
fracsqrtalpha gammad-1 & fracbetad-2 & eta & ddots & vdots \
vdots & vdots & ddots & ddots & fracbetad-2\
fracsqrtalpha gammad-1 & fracbetad-2 & cdots & fracbetad-2 & eta \
endpmatrix$$
where $eta = sqrtleft(1 - beta - fracalphad-1right)left((1 - beta - fracgammad-1right)$
Notice, as it can only help, that $K$ is symmetric, and has strictly positive entries. So far, have only been able to say that the spectral radius lies (up to reordering of the bounds) in
$$left[ sqrt(1 - alpha)(1 - gamma) + sqrtalphagamma; fracsqrtalpha gammad-1+ eta + beta right]$$
since by positivity $rho(K)$ stands in between the minimum and maximum row sums.
Numerically, though, it seems that I have the following dependency $rho(K) = 1 - Oleft( fracf(alpha, gamma)d right)$, and this sort of dependency in $d$ is my goal, but I am a little bit short on tools to confirm it.
linear-algebra eigenvalues-eigenvectors symmetric-matrices upper-lower-bounds spectral-radius
asked Aug 2 at 13:01
ippiki-ookami
303116
Let $alpha, beta, gamma in (0, 1)$ such that $alpha < beta, gamma < beta$. Let $d in mathbbN, d geq 4$.
I am interested in bounding from below the spectral radius $rho(K)$ of the following matrix $K$ of size $d$ (or even better determining it analytically).
$$K = beginpmatrix
sqrt(1 - alpha)(1 - gamma) & fracsqrtalpha gammad-1 & cdots & cdots & fracsqrtalpha gammad-1 \
fracsqrtalpha gammad-1 & eta & fracbetad-2 & cdots & fracbetad-2 \
fracsqrtalpha gammad-1 & fracbetad-2 & eta & ddots & vdots \
vdots & vdots & ddots & ddots & fracbetad-2\
fracsqrtalpha gammad-1 & fracbetad-2 & cdots & fracbetad-2 & eta \
endpmatrix$$
where $eta = sqrtleft(1 - beta - fracalphad-1right)left((1 - beta - fracgammad-1right)$
Notice, as it can only help, that $K$ is symmetric, and has strictly positive entries. So far, have only been able to say that the spectral radius lies (up to reordering of the bounds) in
$$left[ sqrt(1 - alpha)(1 - gamma) + sqrtalphagamma; fracsqrtalpha gammad-1+ eta + beta right]$$
since by positivity $rho(K)$ stands in between the minimum and maximum row sums.
Numerically, though, it seems that I have the following dependency $rho(K) = 1 - Oleft( fracf(alpha, gamma)d right)$, and this sort of dependency in $d$ is my goal, but I am a little bit short on tools to confirm it.
linear-algebra eigenvalues-eigenvectors symmetric-matrices upper-lower-bounds spectral-radius
asked Aug 2 at 13:01
ippiki-ookami
303116
asked Aug 2 at 13:01
ippiki-ookami
303116
asked Aug 2 at 13:01
ippiki-ookami
303116
asked Aug 2 at 13:01
ippiki-ookami
303116
303116
add a comment |Â
add a comment |Â
1 Answer
1
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Let $u=(1,0,ldots,0)^T, v=frac1sqrtd-1(0,1,ldots,1)^T, p=fracsqrtalphagammad-1, q=fracbetad-2$ and $r=sqrt(1-alpha)(1-gamma)$. Then
$$
K=(r-eta+q)uu^T+sqrtd-1,p(uv^T+vu^T)+(d-1)qvv^T+(eta-q)I.
$$
So, $K-(eta-q)I$ has rank $le2$ and its restriction on $operatornamespanu,v$ has a matrix representation
$$
A=pmatrixr-eta+q&sqrtd-1,p\ sqrtd-1,p&(d-1)q.
$$
When $d$ is large, $eta-q>0$ and $A$ is entrywise positive. Hence
beginalign
rho(K)
&=eta-q+rho(A)\
&=eta-q+
frac(r-eta+dq)+sqrtleft[r-eta-(d-2)qright]^2+4(d-1)p^22\
&=frac(r+eta+beta)+sqrt(r-eta-beta)^2+frac4alphagammad-12\
endalign
and it is up to you to find an asymptotic bound for it.
answered Aug 2 at 16:17
user1551
66.7k564122
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $u=(1,0,ldots,0)^T, v=frac1sqrtd-1(0,1,ldots,1)^T, p=fracsqrtalphagammad-1, q=fracbetad-2$ and $r=sqrt(1-alpha)(1-gamma)$. Then
$$
K=(r-eta+q)uu^T+sqrtd-1,p(uv^T+vu^T)+(d-1)qvv^T+(eta-q)I.
$$
So, $K-(eta-q)I$ has rank $le2$ and its restriction on $operatornamespanu,v$ has a matrix representation
$$
A=pmatrixr-eta+q&sqrtd-1,p\ sqrtd-1,p&(d-1)q.
$$
When $d$ is large, $eta-q>0$ and $A$ is entrywise positive. Hence
beginalign
rho(K)
&=eta-q+rho(A)\
&=eta-q+
frac(r-eta+dq)+sqrtleft[r-eta-(d-2)qright]^2+4(d-1)p^22\
&=frac(r+eta+beta)+sqrt(r-eta-beta)^2+frac4alphagammad-12\
endalign
and it is up to you to find an asymptotic bound for it.
answered Aug 2 at 16:17
user1551
66.7k564122
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
add a comment |Â
up vote
1
down vote
accepted
Let $u=(1,0,ldots,0)^T, v=frac1sqrtd-1(0,1,ldots,1)^T, p=fracsqrtalphagammad-1, q=fracbetad-2$ and $r=sqrt(1-alpha)(1-gamma)$. Then
$$
K=(r-eta+q)uu^T+sqrtd-1,p(uv^T+vu^T)+(d-1)qvv^T+(eta-q)I.
$$
So, $K-(eta-q)I$ has rank $le2$ and its restriction on $operatornamespanu,v$ has a matrix representation
$$
A=pmatrixr-eta+q&sqrtd-1,p\ sqrtd-1,p&(d-1)q.
$$
When $d$ is large, $eta-q>0$ and $A$ is entrywise positive. Hence
beginalign
rho(K)
&=eta-q+rho(A)\
&=eta-q+
frac(r-eta+dq)+sqrtleft[r-eta-(d-2)qright]^2+4(d-1)p^22\
&=frac(r+eta+beta)+sqrt(r-eta-beta)^2+frac4alphagammad-12\
endalign
and it is up to you to find an asymptotic bound for it.
answered Aug 2 at 16:17
user1551
66.7k564122
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $u=(1,0,ldots,0)^T, v=frac1sqrtd-1(0,1,ldots,1)^T, p=fracsqrtalphagammad-1, q=fracbetad-2$ and $r=sqrt(1-alpha)(1-gamma)$. Then
$$
K=(r-eta+q)uu^T+sqrtd-1,p(uv^T+vu^T)+(d-1)qvv^T+(eta-q)I.
$$
So, $K-(eta-q)I$ has rank $le2$ and its restriction on $operatornamespanu,v$ has a matrix representation
$$
A=pmatrixr-eta+q&sqrtd-1,p\ sqrtd-1,p&(d-1)q.
$$
When $d$ is large, $eta-q>0$ and $A$ is entrywise positive. Hence
beginalign
rho(K)
&=eta-q+rho(A)\
&=eta-q+
frac(r-eta+dq)+sqrtleft[r-eta-(d-2)qright]^2+4(d-1)p^22\
&=frac(r+eta+beta)+sqrt(r-eta-beta)^2+frac4alphagammad-12\
endalign
and it is up to you to find an asymptotic bound for it.
answered Aug 2 at 16:17
user1551
66.7k564122
Let $u=(1,0,ldots,0)^T, v=frac1sqrtd-1(0,1,ldots,1)^T, p=fracsqrtalphagammad-1, q=fracbetad-2$ and $r=sqrt(1-alpha)(1-gamma)$. Then
$$
K=(r-eta+q)uu^T+sqrtd-1,p(uv^T+vu^T)+(d-1)qvv^T+(eta-q)I.
$$
So, $K-(eta-q)I$ has rank $le2$ and its restriction on $operatornamespanu,v$ has a matrix representation
$$
A=pmatrixr-eta+q&sqrtd-1,p\ sqrtd-1,p&(d-1)q.
$$
When $d$ is large, $eta-q>0$ and $A$ is entrywise positive. Hence
beginalign
rho(K)
&=eta-q+rho(A)\
&=eta-q+
frac(r-eta+dq)+sqrtleft[r-eta-(d-2)qright]^2+4(d-1)p^22\
&=frac(r+eta+beta)+sqrt(r-eta-beta)^2+frac4alphagammad-12\
endalign
and it is up to you to find an asymptotic bound for it.
answered Aug 2 at 16:17
user1551
66.7k564122
edited yesterday
answered Aug 2 at 16:17
user1551
66.7k564122
answered Aug 2 at 16:17
user1551
66.7k564122
answered Aug 2 at 16:17
user1551
66.7k564122
66.7k564122
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
add a comment |Â
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
After computation from your expression, an interesting lower bound then turns out to be $1 - fracalpha + gammad-1$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ?
– ippiki-ookami
yesterday
1
1
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
@ippiki-ookami I should write "rank $le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold.
– user1551
yesterday
add a comment |Â
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