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Let $k$ be a field, and consider the algebraic group $G=GL_n(k)$. For any partition $n_1+n_2+\ldots+n_m=n$, we have a parabolic subgroup of the form

$$P=\begin{pmatrix}A_1&\star&\star&\cdots&\star\\0&A_2&\star&\cdots&\star\\0&0&A_3&\cdots&\star\\\vdots&\vdots&\vdots&\vdots&\vdots\\0&0&0&\cdots&A_m\end{pmatrix}$$

where $A_i\in GL_{n_i}(k)$. I've been told that the unipotent radical of $P$ is

$$U=\begin{pmatrix}I_{n_1}&\star&\star&\cdots&\star\\0&I_{n_2}&\star&\cdots&\star\\0&0&I_{n_3}&\cdots&\star\\\vdots&\vdots&\vdots&\vdots&\vdots\\0&0&0&\cdots&I_{n_m}\end{pmatrix}$$

I have two related questions concerning $U$.

  1. I can see that $U$ is unipotent and solvable, but why is it normal and connected, and why is it the largest such subgroup of $P$?
  2. What is the form of the matrices in $\mathfrak{gl}_n$ that are in the lie algebra of $U$?
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