# Solving Ax=b, when b is an eigenvector

1. Apr 5, 2009

### azdang

Let A be a nonsingular matrix. What can you say about the convergence of GMRES to the solution of Ax=b when b is an eigenvector of A?

I know that if A is a nonsingular matrix with minimal polynomial of degree m, then GMRES solves Ax=b in at most m iterations.

Because b is an eigenvector of A then: (A-$$\lambda$$I)b=0.

The only thing I've sort of come up with is that because b is an eigenvector, it cannot equal 0. At first I though that that implies that A-$$\lambda$$I=0, but I'm thinking that that is not necessarily true. However, IF it is true that A-$$\lambda$$I=0, I completely understand, because this would imply that the minimal polynomial is q(x)=x-$$\lambda$$. This would be a minimal polynomial of degree 1, so GMRES would converge in 1 iteration. Can anyone help me out because I don't think this reasoning is correct, but I'm thinking that A-$$\lambda$$ is not necessarily nonsingular...or is it? Thank you!!

2. Apr 6, 2009

### azdang

Can anyone offer any insight?

3. Apr 7, 2009

### VKint

I just read the Wikipedia page on GMRES. Perhaps I misinterpret, but it seems like GMRES sould solve $$Ax = b$$ in just one step if $$b$$ is an eigenvector, since the exact solution $$x = b/\lambda$$ is in the span of $$b$$ (i.e., the first Krylov subspace).

4. Apr 8, 2009

### azdang

Oh wow! That makes so much sense and seems so obvious. Thank you!