A square matrix with entries from is called a Jordan matrix if for some Show that if then
Remark: Using Jordan normal form the above result is easily extended to any matrix with entries from and with this property that for any eigenvalue of
A square matrix with entries from is called a Jordan matrix if for some Show that if then
Remark: Using Jordan normal form the above result is easily extended to any matrix with entries from and with this property that for any eigenvalue of
We write: where
Now note that M represents the following digraph: 1->2->3->...->n . When we exponentiate M (remember that is the number of paths of lenght m from i to j ) we get: (1) (when I say dimension I am talking about the dimension of our matrices) - From here it is clear that . (2)
Where does this come in handy? Well: ( )
Consider then - by (2) -
Then but, look back at (1) if our entry here is not 0 already, it is only one of the terms of the sum , i.e. for some and this clearly tends to 0 since is fixed ; and
beautiful! now suppose that is a square matrix with entries in let be all distinct eigenvalues of then we have: where is some invertible matrix and
where each is a Jordan matrix with on the diagonal. anywhere else in is 0. therefore:
and thus if for all then this is an important result in applied and pure linear algebra and a key to the solution of skeboy's question in http://www.mathhelpforum.com/math-he...-matrices.html.
skeboy doesn't seem to be looking for a rigorous proof of his question, which, by the way, is a part of the beautiful Perron-Frobenius theorem. i'll post my solution to his problem later.