Definition. Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $\lambda_1$ denoted $K_\lambda$, is the subset of $V$ defined by
$K_\lambda$ = $\{x \in V: (T-\lambda I)^p(x) = 0$ for some positive integer $p\}$.
To show that $K_\lambda$ is $T$-invariant, consider any $x \in K_\lambda$. Choose a positive integer $p$ such that $(T-\lambda)^p = 0$. Then
$(T-\lambda I)^pT(x) = T(T-\lambda I)^p(x) = T(0) = 0$
I wanted to know how we go from $(T-\lambda I)^pT(x) = T(T-\lambda I)^p(x)$?
Since $T$ commutes with both $T$ and $\lambda\operatorname{Id}$, $T$ commutes with $T-\lambda\operatorname{Id}$ and therefore it comutes with any power of $T-\lambda\operatorname{Id}$.