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

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

*The if part:*

Let be the smallest positive integer such that = 0. Then otherwise , which would mean is invertible, a contradiction.

Let us denote by , to save time.

If , then , so . We note that .

Then, a basis for can be extended to a basis for , which can be extended to a basis for and so on, eventually we shall obtain a basis for , or . Since is the smallest integer such that we have whenever Let this basis be where is a basis for .

Let be the matrix of with respect to this basis. Then the first columns of are zero, the next columns have the last entries zero, the next columns have the last entries zero and so on. This is because . Clearly, 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 be upper triangular and be the basis w.r.t which the matrix of is . Then so proceeding this way, we see that , so

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