- #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 ] { x } = λ [ M ] { x }

There is the characteristic matrix which is a function of

[ 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

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

OK, so the eigenproblem matrix EQ is partitioned into a singular boundary section and an internal section for the rest of the coordinates

G

[ G

The value for φ

{ φ

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

G

_{bb}( λ ) φ_{b}+ { G_{bi}( λ ) }^{T}{ φ_{b}} = 0[ G

_{ib}( λ ) ] { φ_{b}} + [ G_{ii}( λ ) ] { φ_{i}} = 0The value for φ

_{b}is then assigned some dummy value (i.e., typically*1*), so that the latter partition EQ becomes{ φ

_{i}} = [ G_{ii[/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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