Answer to Question #214464 in Abstract Algebra for Komal

Question

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

Answer 1

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={x$\epsilon$ V|T^m-1x=0}

W is the null space of T^m-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 x^m and therefore

m$\le$ dim V