Question

In problem set 7, section 6.1, problem 19, how does he so easily state that P(A)x = p(lamda)x? It is not hard to prove for polynomials with positive powers, but it is not so obvious (to me) for the case of 1/(B^2+I).

Answer 1

Professor Strang did not state that. First try to see whether the eigenvalues of B determine the eigenvalues of B^2+I. Then if that works, see if you can find the eigenvalues of the inverse of B^2+I.

Answer 2

Well, in the solution he says "if Ax = lamda*x, then any polynomial p(t) yields p(A)x = p(lamda)*x. Once you accept this quite general rule, then the problem is easy to solve. That first statement was surprising and not something I would find obvious. So I thought this was tricky for a non-challenge problem -- not that I am complaining :)

Answer 3

If Ax = lambda*x, then A^n*x = lambda^n*x. From there it is not hard to see that p(A)x = p(lambda)x.