Question

suppose $N\in L(V)is nilpotent(i.e there exist m\geq 0 s.t N^m=0)$

prove that 0 is the only eigenvalue of N

(hint: fist show 0 is an eigenvalue of N, and then show if $\lambda$ is any eigenvalue then $\lambda$=0

Answer 1

Solne Given: NEL(V) is nilpotent (re.) mzo. ..Nm-o. to prove o is the only eighen value of a Proof: Suppose that has another eigenvalue dt0 so that $(x)=dx. & is an eigen vector corresponding to d) Then, ’lv) = 0 -(OW) = 0 -) 2(w). = n($h-)(x) = ..... = A x 0. we have a contradiction, so o cant another eigen value except o. have Alter A is nilpotent ,KEN ,AK=0 Let & be an eighen value of A and let a be the eigen tatue Vector corresp. to do then, Ax=t& ( satisfy the equality) Å x = NAx = der (multiplying by A) in 11 repeatedly multiply by A. - Ak n = dkre. Now, since Ak=0, we get the nin is zero since x is an eigen vector and hence non zero by definition. we get dk=0 , henced =0 proved vector