# A weird eigenvalue problem

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 $$x$$ ($$b = cx$$).

Birkan

Last edited:

jbunniii
Homework Helper
Gold Member
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 $x$ ($b = cx$).

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

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

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!

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!

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.