Let $E$ be a projection of $V$ and let $T$ be a linear operator on $V$. Prove that the range of $E$ is invariant under $T$ if and only if $ETE=TE$. Prove that both the range and null space of $E$ are invariant under $T$ if and only if $ET = TE$.
We will make use of following lemma: Suppose $E:V\to V$ is projection. Then $\beta \in R_E$$\iff$$E(\beta)=\beta$.
My attempt: $(\Rightarrow)$ Let $\alpha \in V$. Then $E(\alpha)\in R_E$. Since $R_E$ is invariant under $T$, we have $T(E(\alpha))\in R_E$. So $E(T(E(\alpha)))=T(E(\alpha))$. Thus $ETE(\alpha)=TE(\alpha)$, $\forall \alpha \in V$. Hence $ETE=TE$. $(\Leftarrow)$ Let $\alpha \in R_E$. Then $E(\alpha)=\alpha$. Since $ETE=TE$, we have $T(\alpha)=T(E(\alpha))=E(T(E(\alpha)))\in R_E$. Thus $T(\alpha)\in R_E$, $\forall \alpha \in R_E$. Hence $R_E$ is invariant under $T$.
$(\Leftarrow)$ Let $\alpha \in R_E$. Then $E(\alpha)=\alpha$. Since $ET=TE$, we have $T(\alpha)=T(E(\alpha))=E(T(\alpha))\in R_E$. Thus $R_E$ is invariant under $T$. Let $\alpha \in N_E$. Then $E(\alpha)=0$. Since $ET=TE$, we have $0=T(0)=T(E(\alpha))=E(T(\alpha))$. So $T(\alpha)\in N_E$. Thus $N_E$ is invariant under $T$. $(\Rightarrow)$ Since $E:V\to V$ is a projection, we have $V=R_E\oplus N_E$. Let $\alpha \in V$. Then $\alpha =\alpha_1 +\alpha_2$, where $\alpha_1\in R_E$ and $\alpha_2\in N_E$. So $E(\alpha_1)=\alpha_1$, $E(\alpha_2)=0$ and $TE(\alpha)=TE(\alpha_1)+TE(\alpha_2)$. Since $R_E$ is invariant under $T$, we have $T(\alpha_1)\in R_E$. So $T(E(\alpha_1))=T(\alpha_1)=E(T(\alpha_1))$. Since $N_E$ is invariant under $T$, we have $T(\alpha_2)\in N_E$. So $T(E(\alpha_2))=T(0)=0=E(T(\alpha_2))$. Thus $TE(\alpha)=TE(\alpha_1)+TE(\alpha_2)=ET(\alpha_1)+ET(\alpha_2)=ET(\alpha)$, $\forall \alpha \in V$. Hence $TE=ET$. Is my proof correct?
Here is an alternative slick proof of second part.