- #1
swampwiz
- 571
- 83
NOTE: For the answers to all these questions, I'd like an explanation (or a reference to a book or internet page) of how the answer has been derived.
This question can be presumed to be for the general eigenproblem in which [ K ] & [ M ] are Hermitian matrices, with [ M ] also being positive definite, or [ K ] is a normal matrix and [ M ] is the identity matrix. My Question #0 is whether these conditions must be met for there to be a complete eigensolution, or are they more narrow or broad. (I understand that there is finagling that can be done to get [ K ] & [ M ] to get [ M ] to be positive definite, which is called a positive definite pencil)
[ K ] { x } = λ [ M ] { x }
There is the characteristic matrix which is a function of λ
[ G( λ ) ] = [ K ] - λ[ M ]
and the eigenproblem matrix EQ
[ G( λ ) ] { φ } = { 0 }
from which the set of λ is solved by setting the determinant of [ G( λ ) ] to 0. So far so good.
The next step is to solve for the eigenvectors corresponding to each eigenvalue (presume here that the eigenvalues are distinct - the question of what to do if there are repeated values is a question for another thread). Now I get that [ G( λ ) ] has some rank of linear dependency as its determinant is de facto 0 as per the condition. My Question #1 is whether it always has linear dependency of rank 1, or can it have a higher rank.
OK, so the eigenproblem matrix EQ is partitioned into a singular boundary section and an internal section for the rest of the coordinates
Gbb( λ ) φb + { Gbi( λ ) }T { φb } = 0
[ Gib( λ ) ] { φb } + [ Gii( λ ) ] { φi } = 0
The value for φb is then assigned some dummy value (i.e., typically 1), so that the latter partition EQ becomes
{ φi } = [ Gii[/SUB ]( λ ) ]-1 [ Gib( λ ) ]
So obviously [ Gii( λ ) ] must be invertible, and thus b must be chosen to result in this. My Question #2 is whether it is guaranteed that there will always be some b such that the resulting [ Gii( λ ) ] is invertible. My Question #3 is if it turns out that it is not invertible, does that imply that the value of that coordinate's element in { φ } will eventually be calculated to be 0, and if so, is there the converse implication.
And as for Gbb( λ ), there doesn't seem to be the condition that it not be 0 as nothing is being solved for the b partition, although it sure seems like there should be. My Question is #4 is whether there is such a condition, and if so, it is a condition that is somehow always met, and if not, does that mean that that coordinate cannot be chosen for b.
Thanks
NOTE: For the answers to all these questions, I'd like an explanation (or a reference to a book or internet page) of how the answer has been derived.
This question can be presumed to be for the general eigenproblem in which [ K ] & [ M ] are positive Hermitian matrices, or [ K ] is a normal matrix and [ M ] is the identity matrix.
[ K ] { x } = λ [ M ] { x }
There is the characteristic matrix which is a function of λ
[ G( λ ) ] = [ K ] - λ[ M ]
and the eigenproblem matrix EQ
[ G( λ ) ] { φ } = { 0 }
from which the set of λ is solved by setting the determinant of [ G( λ ) ] to 0. So far so good.
The next step is to solve for the eigenvectors corresponding to each eigenvalue (presume here that the eigenvalues are distinct - the question of what to do if there are repeated values is a question for another thread). Now I get that [ G( λ ) ] has some rank of linear dependency as its determinant is de facto 0 as per the condition. My Question #1 is whether it always has linear dependency of rank 1, or can it have a higher rank.
OK, so the eigenproblem matrix EQ is partitioned into a singular boundary section and an internal section for the rest of the coordinates
Gbb( λ ) φb + { Gbi( λ ) }T { φb } = 0
[ Gib( λ ) ] { φb } + [ Gii( λ ) ] { φi } = 0
The value for φb is then assigned some dummy value (i.e., typically 1), so that the latter partition EQ becomes
{ φi } = [ Gii( λ ) ]-1 [ Gib( λ ) ]
So obviously [ Gii( λ ) ] must be invertible, and thus b must be chosen to result in this. My Question #2 is whether it is guaranteed that there will always be some b such that the resulting [ Gii( λ ) ] is invertible. My Question #3 is if it turns out that it is not invertible, does that imply that the value of that coordinate's element in { φ } will eventually be calculated to be 0, and if so, is there the converse implication.
And as for Gbb( λ ), there doesn't seem to be the condition that it not be 0 as nothing is being solved for the b partition, although it sure seems like there should be. My Question is #4 is whether there is such a condition, and if so, it is a condition that is somehow always met, and if not, does that mean that that coordinate cannot be chosen for b.
Thanks
This question can be presumed to be for the general eigenproblem in which [ K ] & [ M ] are Hermitian matrices, with [ M ] also being positive definite, or [ K ] is a normal matrix and [ M ] is the identity matrix. My Question #0 is whether these conditions must be met for there to be a complete eigensolution, or are they more narrow or broad. (I understand that there is finagling that can be done to get [ K ] & [ M ] to get [ M ] to be positive definite, which is called a positive definite pencil)
[ K ] { x } = λ [ M ] { x }
There is the characteristic matrix which is a function of λ
[ G( λ ) ] = [ K ] - λ[ M ]
and the eigenproblem matrix EQ
[ G( λ ) ] { φ } = { 0 }
from which the set of λ is solved by setting the determinant of [ G( λ ) ] to 0. So far so good.
The next step is to solve for the eigenvectors corresponding to each eigenvalue (presume here that the eigenvalues are distinct - the question of what to do if there are repeated values is a question for another thread). Now I get that [ G( λ ) ] has some rank of linear dependency as its determinant is de facto 0 as per the condition. My Question #1 is whether it always has linear dependency of rank 1, or can it have a higher rank.
OK, so the eigenproblem matrix EQ is partitioned into a singular boundary section and an internal section for the rest of the coordinates
Gbb( λ ) φb + { Gbi( λ ) }T { φb } = 0
[ Gib( λ ) ] { φb } + [ Gii( λ ) ] { φi } = 0
The value for φb is then assigned some dummy value (i.e., typically 1), so that the latter partition EQ becomes
{ φi } = [ Gii[/SUB ]( λ ) ]-1 [ Gib( λ ) ]
So obviously [ Gii( λ ) ] must be invertible, and thus b must be chosen to result in this. My Question #2 is whether it is guaranteed that there will always be some b such that the resulting [ Gii( λ ) ] is invertible. My Question #3 is if it turns out that it is not invertible, does that imply that the value of that coordinate's element in { φ } will eventually be calculated to be 0, and if so, is there the converse implication.
And as for Gbb( λ ), there doesn't seem to be the condition that it not be 0 as nothing is being solved for the b partition, although it sure seems like there should be. My Question is #4 is whether there is such a condition, and if so, it is a condition that is somehow always met, and if not, does that mean that that coordinate cannot be chosen for b.
Thanks
NOTE: For the answers to all these questions, I'd like an explanation (or a reference to a book or internet page) of how the answer has been derived.
This question can be presumed to be for the general eigenproblem in which [ K ] & [ M ] are positive Hermitian matrices, or [ K ] is a normal matrix and [ M ] is the identity matrix.
[ K ] { x } = λ [ M ] { x }
There is the characteristic matrix which is a function of λ
[ G( λ ) ] = [ K ] - λ[ M ]
and the eigenproblem matrix EQ
[ G( λ ) ] { φ } = { 0 }
from which the set of λ is solved by setting the determinant of [ G( λ ) ] to 0. So far so good.
The next step is to solve for the eigenvectors corresponding to each eigenvalue (presume here that the eigenvalues are distinct - the question of what to do if there are repeated values is a question for another thread). Now I get that [ G( λ ) ] has some rank of linear dependency as its determinant is de facto 0 as per the condition. My Question #1 is whether it always has linear dependency of rank 1, or can it have a higher rank.
OK, so the eigenproblem matrix EQ is partitioned into a singular boundary section and an internal section for the rest of the coordinates
Gbb( λ ) φb + { Gbi( λ ) }T { φb } = 0
[ Gib( λ ) ] { φb } + [ Gii( λ ) ] { φi } = 0
The value for φb is then assigned some dummy value (i.e., typically 1), so that the latter partition EQ becomes
{ φi } = [ Gii( λ ) ]-1 [ Gib( λ ) ]
So obviously [ Gii( λ ) ] must be invertible, and thus b must be chosen to result in this. My Question #2 is whether it is guaranteed that there will always be some b such that the resulting [ Gii( λ ) ] is invertible. My Question #3 is if it turns out that it is not invertible, does that imply that the value of that coordinate's element in { φ } will eventually be calculated to be 0, and if so, is there the converse implication.
And as for Gbb( λ ), there doesn't seem to be the condition that it not be 0 as nothing is being solved for the b partition, although it sure seems like there should be. My Question is #4 is whether there is such a condition, and if so, it is a condition that is somehow always met, and if not, does that mean that that coordinate cannot be chosen for b.
Thanks
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