So, if B is diagonalizable (over the complex numbers), then Bp -> 0 is equivalent to all the eigenvalues being norm less than one.
If B isn't diagonalizable, you'd have to look at the Jordan normal form. I think if all the eigenvalues are norm less than 1, it still goes to zero. But I'm having trouble working out the details in my head, and I'm too lazy right now to actually get a piece of paper and do the calculation out.
Consider a Jordan block with eigenvalue lambda, where norm(lambda) < 1.
If lambda = 0, then powers of the block will keep becoming more upper triangular with each power, so after n steps it will just be 0.
If lambda != 0, the entries k places above the diagonal will never be more ((k+1)/lambda) times bigger than the entries just below them. (This can be proven by induction on k.) This means that no entry will ever be more than (n!)/(lambda^n) times as big as the diagonal. Since this is a constant, and the diagonal goes to zero, all entries will go to zero.
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(no subject)
Date: 2007-11-12 01:21 am (UTC)If B isn't diagonalizable, you'd have to look at the Jordan normal form. I think if all the eigenvalues are norm less than 1, it still goes to zero. But I'm having trouble working out the details in my head, and I'm too lazy right now to actually get a piece of paper and do the calculation out.
(no subject)
Date: 2007-11-12 09:52 pm (UTC)If lambda = 0, then powers of the block will keep becoming more upper triangular with each power, so after n steps it will just be 0.
If lambda != 0, the entries k places above the diagonal will never be more
((k+1)/lambda) times bigger than the entries just below them. (This can be proven by induction on k.) This means that no entry will ever be more than (n!)/(lambda^n) times as big as the diagonal. Since this is a constant, and the diagonal goes to zero, all entries will go to zero.
(no subject)
Date: 2007-11-12 10:21 pm (UTC)(no subject)
Date: 2007-11-12 10:52 pm (UTC)