Can Ax = λx + b Be Solved in Euclidean Spaces?

[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

 

[jbunniii]

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

(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!

 

[ahmethungari]

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.