# 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 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 $T^n=0.$