If G is a linear group, then G has the largest normal subgroup consisting of unipotent matrices. This is called the unipotent radical of G. If G is brought by conjugation in Mn(K) to a block diagonal form with irreducible diagonal blocks, then the unipotent radical of G is the kernel of the projection onto the diagonal blocks. How to understand "If G is brought by conjugation in Mn(K) to a block diagonal form with irreducible diagonal blocks, then the unipotent radical of G is the kernel of the projection onto the diagonal blocks"? What does projection mean? Thanks.

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You can replace "brought by conjugation to" with "conjugate to" : Up to conjugacy, you can write $G$ as a group of block upper triangular, that is of the form
$\left( \begin{matrix} A_1 & * & \ldots \\ & \ddots &*\\ 0& & A_r \end{matrix} \right)$
such that the blocks $A_i$ on the diagonal are irreducible (meaning you cannot get smaller blocks if you conjugate). You get such a decomposition by looking at the action of $G \subset \textrm{GL}_n$ on $K^n$ (you need a maximal sequence $\{0\} \subsetneq V_1 \subsetneq \ldots \subsetneq V_r = K^n$ of $G$-stable subspaces).
The unipotent radical does NOT contain all unipotent matrices. It's the biggest NORMAL subgroup of $G$ consisting only of unipotents elements. Take for example, $G = \textrm{GL}_n$ or $\textrm{SL}_n$, then it's irreducible (it's not conjugate to a smaller block upper triangular matrices subgroup). So the decomposition reduces to just one block ($G$ itself). The result just states that's there is no (non trivial) normal subgroup of unipotent matrices although there are plenty of unipotent matrices (we say $G$ is reductive). In the general case, the blocks that appear are also reductive. – Joel Cohen May 11 '11 at 2:59