A weird eigenvalue problem

[ahmethungari]

Hi,

 

Is there any solution for the following problem:

 

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

 

Here [math]x[/math] seems to be an eigenvector of [math]A[/math] but with an extra translation vector [math]b[/math].

I cannot say whether [math]b[/math] is parallel to [math]x \quad[/math], ([math]b = cx[/math]).

 

Thank you in advance for your help...

 

Birkan

Edited May 25, 2009 by ahmethungari

[Bignose]

This forum doesn't support using $ as the math tags like regular LaTeX. You need to use (math) and (/math) but with square brackets in place of the beginning and ending $'s and the LaTeX will appear correctly.

[Severian]

If [math]A- \lambda \mathbf{1}[/math] has an inverse, then the solution is [math]x= \left[ A- \lambda \mathbf{1} \right]^{-1} b[/math]

[ahmethungari]

If [math]A- \lambda \mathbf{1}[/math] has an inverse, then the solution is [math]x= \left[ A- \lambda \mathbf{1} \right]^{-1} b[/math]

 

The problem is we do not know the [math]\lambda[/math] either. Only [math]A[/math] and [math]b[/math] are knowns.

[Severian]

Then you don't have enough information. If x is an N-dimensional vector, you have N+1 unknowns with only N equations. In a traditional eigenvalue problem, this doesn't matter because you don't need the magnitude of x (since every term is linear in x), but here the b term sets a scale.