I was reading the wikipedia page the other day, and it says that Schur's decomposition implies the existence of a sequence of a invariant ( ie. An operator invariant subspaces . I just need a small hint to see that the fact is true .
Schur's decomposition and sequence of invariant subspaces .
0
$\begingroup$
linear-algebra
-
0@KarolisJuodelė : Ah , its just taking each basis and keep on adding one more to form a sequence right ? – 2012-12-27
1 Answers
0
If $\{ \vec{v}_1 \dots \vec{v}_n \}$ is the orthogonal basis then $\{0\} \subset \text{span}\{\vec{v}_1\} \subset \text{span} \{ \vec{v}_1, \vec{v}_2 \} \subset \dots \subset \text{span}\{ \vec{v}_1 \dots \vec{v}_n \} = \mathbb{C}^n$ is the nested sequence of invariant subspaces. This sequence follows quite naturally from the construction outlined in Proof section.