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Let A be nxn and nilpotent, that is A^m = 0 for some m>/=1.?
Let A be nxn and nilpotent, that is A^m = 0 for some m>/=1.
a)if k is an eigenvalue of A, show that k=0.
b)Show that the characteristic polynomial c_A(x) of A is given by c_A(x) = x^n.
1 Answer
- kbLv 79 years agoFavorite Answer
a) Suppose that Av = λv for some scalar λ and nonzero vector v.
Then, A^2 v = A(Av) = A(λv) = λ (Av) = λ(λv) = λ^2 v.
Inductively, we have A^k v = λ^k v.
Letting k = m yields λ^k * v = A^k v = 0v = 0.
Hence, λ^k = 0 and thus λ = 0.
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b) Since λ = 0 is the only eigenvalue, and A is n x n, we see that it must equal (x - 0)^n = x^n.
I hope this helps!
