Every reducible unitary representation is fully reducible ↩

Metadata 
aliases: []
shorthands: {}
created: 2021-10-19 14:15:10
modified: 2022-01-10 04:13:04

Theorem: Every reducible unitary representation is fully reducible

In other words: being unitary makes a reducible representation fully decomposable into irreducible atomic parts.

Proof

group, unitary representation with being a Hermitian product space.

  1. is a nontrivial invariant subspace of , so .
  2. Let be the orthogonal complement of :

So now: and .
Now the question is: Is an invariant subspace of as well?

  1. Choose a random , then because of unitarity.
  2. Then wich means that under any transformation, remains perpendicular to , so it is also an invariant subspace of .
  3. Now we know that has at least two invariant subspaces can be globally blockdiagonalized with block sizes and is decomposable.