# Contraction Mapping and Diagonally Dominant Jacobian

• Aug 14th 2011, 10:20 PM
tsekh
Contraction Mapping and Diagonally Dominant Jacobian
Hello

I'm trying to solve a fixed point problem x=F(x), where F: R^n -> R^n, and would want to show that F is contraction so that the fixed point is unique.

I have seen people using a result which roughly says that if the Jacobian matrix J of F has dominant diagonal, then F is a contraction. Can someone point me to any book/article which has a formal statement and proof of this claim? My statement here may be a little off, and that's why I'm looking for a formal statement/proof to make sure that this is a result that I can use.

I know this result is related to another result that if the spectral radius of J is less than 1, then F is a contraction, if that helps. Thanks.
• Aug 15th 2011, 05:10 AM
Ackbeet
Re: Contraction Mapping and Diagonally Dominant Jacobian
Are the entries of F constant or variable? If constant, then I would simply say that F is a contraction mapping if all its eigenvalues are less than one in magnitude. In general, I think you're looking at a functional analytic argument. Kreyzsig's Introduction to Functional Analysis is excellent.
• Aug 15th 2011, 05:19 AM
tsekh
Re: Contraction Mapping and Diagonally Dominant Jacobian
F is not a constant matrix, it is a function of x.

I know there are numbers of way to verify whether F at x is locally contractive by checking eigenvalues. But eigenvalues are hard to check in application, but I've already verified that the Jacobian matrix of F has dominant diagonal for all (relevant) x. I've seen articles that vaguely said "J is diagonally dominant, so F is a contraction", but without seeing a formal statement and proof I don't feel comfortable using this result.