Let $P$ be a parabolic subgroup of an algebraic group $G$. How to prove that $P = L_P U_P$? Here $L_P$ is the Levi of $P$ and $U_P$ is the unipotent radical of $P$. Thank you very much.
Edit: I think that if $$ P=\left( \begin{matrix} * & * & * & * \\ * & * & * & * \\ 0 & 0 & * & * \\ 0 & 0 & * & * \end{matrix} \right), $$ then $$ U_P=\left( \begin{matrix} 1 & 0 & * & * \\ 0 & 1 & * & * \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right), $$
$$ L_P=\left( \begin{matrix} * & * & 0 & 0 \\ * & * & 0 & 0 \\ 0 & 0 & * & * \\ 0 & 0 & * & * \end{matrix} \right). $$ Is this true? Thank you very much.
The proof just uses the root subgroup structure of $G$ and the fact that up to conjugation every parabolic is a standard parabolic, so we need only prove this when $P = P_I$ for some set $I$ of simple roots.
If $R$ is the root system and $I \subseteq \Delta$ is some subset of simple roots then let $R_I = R \cap \mathbb ZI$. Let $T$ be the torus and for any root $\alpha \in R$ let $U_\alpha$ be the root subgroup corresponding to $\alpha$. Now $P_I$ is generated by $T$ and root subgroups $U_\alpha$ where $\alpha \in R \setminus R_I^-$. $U_I$ is generated by the root subgroups $U_\alpha$ where $\alpha \in R_I^+$ and $L_I$ is generated by $U_\alpha$ where $\alpha \in R \setminus R_I$.
You need to show that $U_I$ is the unipotent radical of $P_I$ and that $L_I$ is reductive, but once you have that you just observe that $L_IU_I$ is generated by $T$ and root subgroups $U_\alpha$ where $\alpha$ is in $$R_I^+ \cup (R \setminus R_I) = R \setminus R_I^-.$$ But that's what generates $P_I$ so $L_IU_I = P_I$.