Question

Who can tell me all parabolic subgroup of GL(3,K) and their Levi decomposition?

Answer 1

This is an elementary question better answered by consulting a textbook. Or start with the Springer encylopedia entry here. What Bruce describes are the isomorphism types of parabolics or Levi subgroups in general linear groups, say over an algebraically closed field $K$ to avoid considering fields of definition. But there are $2^{n-1}$ conjugacy classes of parabolic subgroups in $GL(n,K)$ or $SL(n,K)$, where $n-1$ is the semisimple Lie rank of the group. For $n=3$, the partition $(2,1)$ actually corresponds to two isomorphic but non-conjugate parabolic subgroups or their Levi factors.

Answer 2

The group $GL(n)$ is reductive but not semisimple. Usually one would take the simple group $SL(n)$. Also I am taking the question to mean parabolic subgroups up to conjugacy. According to one definition the parabolic subgroups are the subgroups which contain the lower triangular invertible matrices.These are indexed by partitions of $n$. The Levi subgroup is a product of $GL(k_i)$ where the partition is $(k_1,k_2,\ldots)$.

For n=3 we have the partitions $(1,1,1)$, $(2,1)$, $(3)$. For the partition $(1,1,1)$ the Levi subgroup is the group of diagonal matrices. For $(3)$ the parabolic subgroup and the Levi subgroup are both $GL(3)$. For $(2,1)$ the Levi subgroup is $GL(2)\times GL(1)$.