- #1
azdang
- 80
- 0
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!
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!
