If $N(L-\lambda I)=\text{span}\{ u_0\}$, then $u_0\notin R(L-\lambda I) \Leftrightarrow N((L-\lambda I)^2)=N(L-\lambda I)$?

Question

If $N(L-\lambda I)=\text{span}\{ u_0\}$, then $u_0\notin R(L-\lambda I) \Leftrightarrow N((L-\lambda I)^2)=N(L-\lambda I)$ ?

In Remark 2.6.(p331) of the book "Functional Analysis" written by Rudin, the following statement is found

$u_0\notin R(L-\lambda I)$ means simply that $N((L-\lambda I)^2)=N(L-\lambda I)$

where $u_0$ is used for $N(L-\lambda I)=\text{span}\{ u_0\}$. How do you get this claim? It should be a simple transformation, but I'm having trouble figuring it out.

Answer 1

It's easier to show the contrapositive: if $N(L - \lambda I)^2 \neq N(L - \lambda I)$, then it must hold that $u_0 \in R(L - \lambda I)$.

We can show this as follows. In general, $N(L - \lambda I)^2 \subset N(L - \lambda I)$, so if $N(L - \lambda I)^2 \neq N(L - \lambda I)$, then there exists an $x \in N(L - \lambda I)^2 \setminus N(L - \lambda I)$. That is, $(L - \lambda I)x\neq 0$, but $(L - \lambda I)^2x = 0$. With that, conclude that $(L - \lambda I)x$ is a non-zero element of $N(L - \lambda I)$, which means that $(L - \lambda I)x = ku_0$ for some $k \neq 0$.

Thus, we conclude that $u_0 = (L - \lambda I)(\frac xk)$ is an element of $R(L - \lambda I)$.