# A weird eigenvalue problem

1. May 25, 2009

### ahmethungari

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

Last edited: May 26, 2009
2. May 26, 2009

### jbunniii

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

$$(A - \lambda I)x = b$$

If $$b = 0$$, 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 $$b \neq 0$$, then this is equivalent to

$$b \in image(A - \lambda I)$$

For this to happen, it suffices that $$A - \lambda I$$ be surjective. This is true if and only if $$A - \lambda I$$ 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. May 26, 2009

### ahmethungari

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

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