Show that if A is nilpotent, then the only eigenvalue of A is 0.
Here's my work
I thought about trying to find the characteristic polynomial but I've got too little information to get far...
first off, by supposition you say that λ is non-zero. eigenvectors are by definition non-zero. so you have on the one hand A^n(v) = λ^nv where neither is 0, so A^n(v) is non-zero.
on the other hand A^n(v) = 0v = 0, since A is nilpotent. how can a 0-vector be equal to a non-zero vector? it can't. therefore, λ can't be non-zero (that was our ONLY assumption).
@Fernando: i see that the mimimal polynomial of A has to be of the form x^s (because the only divisors of x^k are of that form), but how do you conclude that the characteristic polynomial has no other roots? unless what you're actually saying is that the minimal polynomial of A actually has to be x^k (which i think it does since A^s , s < k, should not be 0, by the definition of the order k of A). if so, then we can take any polynomial p(A) that A satisfies (including the characteristic polynomial), and knock off the powers of A greater than k to get a polynomial of degree at most k-1 that A also satisfies, and since x^k is minimal, the left-over part must be 0. but then i feel i'm missing something, why can't the characteristic polynomial be of the form: (x^k)^mq(x)? unless you're making a tacit appeal to the fact that q(x) has to split in some extension of the base field (the original poster doesn't indicate what the entries of A actually are, presumably real or complex numbers, but maybe not), so that considered as a matrix in Mn(E) (instead of Mn(F)) q(x) has roots in E, and thus A has eigenvalues in E corresponding to these roots), contradicting the minimality of x^k for A. if so, it seems the long way around. tell me what i'm missing....