Question #214464

If T is nilpotent transformation on V of index m. Prove that m cannot exceed the dimension of V


1
Expert's answer
2021-07-07T16:05:27-0400

Question

If T is nilpotent transformation on V of index m. Prove that m cannot exceed the dimension of V.

Solution;

If T is a linear transformation on V of order m, then dim V \ge m by induction.

Assume the claim herein to be true for index m-1, and let W={xϵ\epsilon V|Tm-1x=0}

W is the null space of Tm-1

Clearly,T maps W onto itself and T is nilpotent of order m-1.

By induction W\ge m-1 since W is a proper subspace of V.

We then conclude that ,the minimal polynomial of T is xm and therefore

m\le dim V







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