Suppose T in L(C^3) is defined by T(Z1,Z2,Z3)=(Z2,Z3,0). Prove that T has no square root.?

More precisely, Prove that there does not exist S in L(C^3) such that S^2=T.

Suppose T in L(C^3) is defined by T(Z1,Z2,Z3)=(Z2,Z3,0). Prove that T has no square root.?
Suppose there was such an S. Then:





S^2 (Z1,Z2,Z3) = (Z2,Z3,0)





which implies S^6 (Z1,Z2,Z3) = (0,0,0), or in other words S^6 = 0. Therefore the minimal polynomial of S must divide x^6, and so must be either x^3, x^2, or x (since it's degree is at most 3 by that theorem that every matrix satisfies its own characteristic polynomial). Thus it must be the case that S^3 = 0, and so S^4 = 0 as well. However S^4 = T^2, which is not zero so this is a contradiction.