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Matrix product and eigen values

1

0

$begingroup$

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

share|cite|improve this question

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

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  $begingroup$
  If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = beginbmatrix0&0\0&0endbmatrix qquad L_2 = beginbmatrix1&-1\-1&1endbmatrix qquad B=beginbmatrix-1 &0 \ 0 & 1endbmatrix$$ 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
  
  

  
  
  
  


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  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
  

  
  
  
  


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  @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
  

  
  
  
  


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  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
  

  
  
  
  


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  $begingroup$
  (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

1

0

$begingroup$

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

share|cite|improve this question

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you count geometric multiplicities, eigenvalues may disappear. Does that count as "increasing"? For example if $$ L_1 = beginbmatrix0&0\0&0endbmatrix qquad L_2 = beginbmatrix1&-1\-1&1endbmatrix qquad B=beginbmatrix-1 &0 \ 0 & 1endbmatrix$$ 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

share|cite|improve this question

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

$endgroup$

share|cite|improve this question

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

share|cite|improve this question

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255

asked Mar 26 at 10:37

Abhiram V PAbhiram V P

255