Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$ with corresponding eigenspace and generalized eigenspace $E_{\lambda}$ and $K_{\lambda}$, respectively. Let $U$ be an invertible operator on $V$ that commutes with $T$ (i.e. $TU = UT$). Prove that $U(E_{\lambda}) = E_{\lambda}$ and $U(K_\lambda) = K_{\lambda}$.
I need to solve the question given above; however, I am unsure of how to proceed exactly; am I required to use the invertibility of $U$ to somehow show the two equalities? Any help would be appreciated!
Hint: Note that $x \in E_\lambda$ if and only if $(T - \lambda I) x = 0$ (where $I$ denotes the identity operator). Show that if $x \in E_\lambda$, we also have $(T - \lambda I) Ux = 0$.
In the same vein, note that $x \in K_\lambda$ if and only if there is a positive integer $k$ for which $(T - \lambda I)^k x = 0$