Centre of the general linear group
Let
Proof
It is clear that scalar transformations form a group isomorphic to
and that . Now let and assume there exists a nonzero such that not an eigenvector of . Then is linearly independent from , and there exists some vector basis with and . Let and . Define linear maps such that in the basis and
. Then implying
, a contradiction. Hence every must be an eigenvector of , so .