Matrix product and eigen values Announcing the arrival of Valued Associate #679: Cesar...
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Matrix product and eigen values
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Eigenvalues of product of a matrix and a diagonal matrixEigenvalues of product of a matrix and a diagonal matrixRelation between trace and Ky Fan normsymmetric normalized Graph Laplacian and symmetric normalized Adjacency matrix eigenvaluesWeighted undirected graphs, complex Laplacian, complex eigenvalues & spectral cluseringWhat is the multiplicity of the largest eigenvalue of a graph?What's the relationship between the rank and eigenvalues of symmetric positive semidefinite matrix (real domain)?Laplacian spectrum of directed network (digraph) and its complementMaximizing the smallest positive eigenvalue of the Laplacian matrix via SDPIs the off-diagonal part of a covariance matrix, $M = Sigma -operatorname{ diag}(Sigma)$ studied?Why are we interested in finding the spectrum of products of graphs?
$begingroup$
Is there any relationship between eigenvalues(or spectrum) of graph Laplacian matrix and the eigenvalues of the product of a real symmetric matrix and the Laplacian matrix?
My problem at hand is as follows :
Let A=L*B.
What is the relationship between spectrum (or eigenvalues) of L with the spectrum of A?
L is Laplacian of an undirected graph, hence real symmetric and singular.
B is a real symmetric matrix.
I want to show that if I increase the magnitude of eigenvalues of L, the eigenvalues of A will also increase. However, all I could find was a trace inequality relationship, and inequality doesn't necessarily lead to any conclusion.
matrices graph-theory eigenvalues-eigenvectors trace graph-laplacian
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
$endgroup$
|
show 7 more comments
$begingroup$
Is there any relationship between eigenvalues(or spectrum) of graph Laplacian matrix and the eigenvalues of the product of a real symmetric matrix and the Laplacian matrix?
My problem at hand is as follows :
Let A=L*B.
What is the relationship between spectrum (or eigenvalues) of L with the spectrum of A?
L is Laplacian of an undirected graph, hence real symmetric and singular.
B is a real symmetric matrix.
I want to show that if I increase the magnitude of eigenvalues of L, the eigenvalues of A will also increase. However, all I could find was a trace inequality relationship, and inequality doesn't necessarily lead to any conclusion.
matrices graph-theory eigenvalues-eigenvectors trace graph-laplacian
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
$endgroup$
$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44
|
show 7 more comments
$begingroup$
Is there any relationship between eigenvalues(or spectrum) of graph Laplacian matrix and the eigenvalues of the product of a real symmetric matrix and the Laplacian matrix?
My problem at hand is as follows :
Let A=L*B.
What is the relationship between spectrum (or eigenvalues) of L with the spectrum of A?
L is Laplacian of an undirected graph, hence real symmetric and singular.
B is a real symmetric matrix.
I want to show that if I increase the magnitude of eigenvalues of L, the eigenvalues of A will also increase. However, all I could find was a trace inequality relationship, and inequality doesn't necessarily lead to any conclusion.
matrices graph-theory eigenvalues-eigenvectors trace graph-laplacian
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
$endgroup$
Is there any relationship between eigenvalues(or spectrum) of graph Laplacian matrix and the eigenvalues of the product of a real symmetric matrix and the Laplacian matrix?
My problem at hand is as follows :
Let A=L*B.
What is the relationship between spectrum (or eigenvalues) of L with the spectrum of A?
L is Laplacian of an undirected graph, hence real symmetric and singular.
B is a real symmetric matrix.
I want to show that if I increase the magnitude of eigenvalues of L, the eigenvalues of A will also increase. However, all I could find was a trace inequality relationship, and inequality doesn't necessarily lead to any conclusion.
matrices graph-theory eigenvalues-eigenvectors trace graph-laplacian
matrices graph-theory eigenvalues-eigenvectors trace graph-laplacian
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
asked Mar 26 at 10:37
Abhiram V PAbhiram V P
166
166
$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44
|
show 7 more comments
$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44
$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44
|
show 7 more comments
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$begingroup$
If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = begin{bmatrix}0&0\0&0end{bmatrix} qquad L_2 = begin{bmatrix}1&-1\-1&1end{bmatrix} qquad B=begin{bmatrix}-1 &0 \ 0 & 1end{bmatrix}$$ then going from $L_1$ to $L_2$ changes the eigenvalues from $0$ and $0$ to $0$ and $2$, certainly as good an increase as we can hope for given that the $L$s are always singular. But the eigenvalues of $L_1B$ are $0$ and $0$ whereas $L_2B$ only has a single $0$. Is that an increase?
$endgroup$
– Henning Makholm
Mar 26 at 11:17
$begingroup$
The topmost question in the "Related" list looks pretty relevant, though it is more picky about the choice for $B$.
$endgroup$
– Henning Makholm
Mar 26 at 11:26
$begingroup$
@HenningMakholm Thank you for the comment. Regarding your first question, I could not understand the difference between the two in "eigenvalues of L1B are 0 and 0 whereas L2B only has a single 0". And were you trying to give a counterexample for the statement? If B is an identity matrix, then L2B would have eigenvalues 0 and 2. I am looking for an analytical proof which says that by increasing eigenvalue of one matrix, the eigenvalue of the product is also increased. I am getting that trend in my case while using the values, but an analytical proof is what I am after.
$endgroup$
– Abhiram V P
Mar 26 at 12:38
$begingroup$
I am asking you whether this counts as a counterexample for you, since $L_1B$ has the eigenvalue $0$ with geometric multiplicity $2$ and $L_2B$ has the eigenvalue $0$ with geometric multiplicity $1$ and no other eigenvalues. One of the eigenvalues of $L_1B$ goes from being $0$ to not being there at all! Do you consider that to be an "increase" or not?
$endgroup$
– Henning Makholm
Mar 26 at 12:41
$begingroup$
(And I don't understand why you say "If $B$ is an identity matrix", since I have explicitly specified what $B$ in my example is -- and that is not the identity matrix").
$endgroup$
– Henning Makholm
Mar 26 at 12:44