Algebra (Artin) Ch.4 Exercise 6.8

Problem statement: A linear operator is nilpotent if some positive power T^k is zero. Prove that T is nilpotent if and only if there is a basis of V such that the matrix of T is upper triangular, with diagonal entries zero.

Solution: (assume that V is finite-dimensional) Let the matrix of T be A .

The if part:

Let k be the smallest positive integer such that T^k = 0. Then ker T \neq {0}, otherwise dim V=dim (Im T) , which would mean A is invertible, a contradiction.

Let us denote ker T^i by K_i , to save time.

If T^i(v)=0, then T^{i+1}(v)=0 , so K_i \subset K_{i+1} . We note that K_k=V .

Then, a basis for K_1 can be extended to a basis for K_2, which can be extended to a basis for K_3 and so on, eventually we shall obtain a basis for K_k , or V . Since k is the smallest integer such that T^k=0, we have K_i \neq K_j whenever i \neq j, 0<i,j<k. Let this basis be (u_11,..,u_1r_1,u_21,..,u_2r_2,...,u_k1,..,u_kr_k), where  (u_11,..,u_ir_i) is a basis for K_i .

Let M be the matrix of T with respect to this basis. Then the first r_1 columns of M are zero, the next r_2 columns have the last n-r_1 entries zero, the next r_3 columns have the last n - (r_1+r_2) entries zero and so on. This is because T(K_{i+1}) \subset K_i . Clearly, M is an upper triangular matrix. We would get a lower triangular matrix if we wrote the basis vectors in the opposite order.

Only if part:

Let A be upper triangular and (v_1,..,v_n) be the basis w.r.t which the matrix of T is A. Then T(v_1)=0, T(v_2)=a_12 v_1, so T^2(v_2)=0, proceeding this way, we see that T^i(v_i)=0, i=1,2,..,n, so T^n=0.

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