Compactness nilpotent operator

Question

Let $A \in L(X) $, where $X$ is Banach space, and given $A^2 = 0$ then A is compact operator.

I found out spectrum of $A$ is zero, but I still can't understand, what should I do next.

Can someone help me with this task?

Answer 1

The statement is false. As a counterexample, consider the following operators over $\ell^2(\Bbb C)$: $$ P(x_1,x_2,x_3,x_4,x_5,\dots) = (0,x_2,0,x_4,0,\dots)\\ R(x_1,x_2,x_3,x_4,\dots) = (0,x_1,x_2,x_3,\dots)\\ A(x_1,x_2,x_3,x_4,\dots) = (R \circ P)(x_1,x_2,x_3,x_4,\dots) = (0,0,x_2,0,x_4,\dots). $$ Verify that $A$ satisfies $A^2 = 0$. On the other hand, we can see that $A$ fails to be compact. One way to see that $A$ is not compact is to note that $A^* \circ A$ is a projection of infinite rank. Indeed, we have $$ \begin{align} A^* \circ A &= (R \circ P)^* \circ (R \circ P) = P^* \circ R^* \circ R \circ P \\ & = P^* \circ (R^* \circ R) \circ P = P \circ \operatorname{id}_X \circ P = P \circ P = P. \end{align} $$