Solving Ax=b, when b is an eigenvector

[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-\lambdaI)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-\lambdaI=0, but I'm thinking that that is not necessarily true. However, IF it is true that A-\lambdaI=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!

 

[azdang]

Can anyone offer any insight?

 

[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).

 

[azdang]

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