A weird eigenvalue problem

  • #1

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Hi,

Is there any solution for the following problem:

[tex]Ax = \lambda x + b[/tex]

Here [tex]x[/tex] seems to be an eigenvector of [tex]A[/tex] but with an extra translation vector [tex]b[/tex].
I cannot say whether [tex]b[/tex] is parallel to [tex]x[/tex] ([tex]b = cx[/tex]).

Thank you in advance for your help...

Birkan
 
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  • #2
jbunniii
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Hi,

Is there any solution for the following problem:

$Ax = \lambda x + b$

Here $x$ seems to be an eigenvector of $A$ but with an extra translation vector $b$.
I cannot say whether $b$ is parallel to $x$ ($b = cx$).

Thank you in advance for your help...

Birkan
You didn't say what vector space you are working with, so I will assume a complex finite-dimensional vector space.

We can rewrite your problem as

[tex](A - \lambda I)x = b[/tex]

If [tex]b = 0[/tex], this has a solution if and only if lambda is an eigenvalue of A. Every map in a complex finite-dimensional vector space has an eigenvalue, so a solution exists in this case.

If [tex]b \neq 0[/tex], then this is equivalent to

[tex]b \in image(A - \lambda I)[/tex]

For this to happen, it suffices that [tex]A - \lambda I[/tex] be surjective. This is true if and only if [tex]A - \lambda I[/tex] is invertible, which is true if and only if lambda is NOT an eigenvalue of A. Thus there are plenty of solutions in this case!
 
  • #3
You didn't say what vector space you are working with, so I will assume a complex finite-dimensional vector space.

We can rewrite your problem as

[tex](A - \lambda I)x = b[/tex]

If [tex]b = 0[/tex], this has a solution if and only if lambda is an eigenvalue of A. Every map in a complex finite-dimensional vector space has an eigenvalue, so a solution exists in this case.

If [tex]b \neq 0[/tex], then this is equivalent to

[tex]b \in image(A - \lambda I)[/tex]

For this to happen, it suffices that [tex]A - \lambda I[/tex] be surjective. This is true if and only if [tex]A - \lambda I[/tex] is invertible, which is true if and only if lambda is NOT an eigenvalue of A. Thus there are plenty of solutions in this case!

I got it. By the way, vector space is actually finite-dimensional (d=9000) Euclidean Space.

Since I do not know the [tex]\lambda[/tex], (only [tex]A[/tex] and [tex]b[/tex] are known) how can I find an numeric solution for that? Is there any way using eigenvalue logic here?
Such as
-- find eigenvalues of [tex]A[/tex],
-- check if [tex]b[/tex] is parallel to any
-- select the appropriate eigenvector etc.
 

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