For Banach spaces, the first example of an operator without an invariant subspace was constructed by Enflo . ( For Hilbert spaces, the invariant subspace problem remains open .)
32.
We consider the representation \ rho ( A ) = A for all A \ in G . The subspace \ C e _ 1 is a G-invariant subspace.
33.
One implication here is trivial, and the other, starting from a finite-dimensional invariant subspace, follows from complete reducibility of representations of " T ".
34.
It is not known that, in general, whether a bounded operator " A " on a Hilbert space " H " has a nontrivial invariant subspace.
35.
:: : : : The relation of Burnside's lemma to invariant subspaces has been known for decades, and is discussed in Lyubich's book, previously cited.
36.
An invariant subspace of dimension 1 will be acted on by " T " by a scalar, and consists of invariant vectors if and only if that scalar is 1.
37.
Conversely, diagonalizable operators are easily seen to be semi-simple, as invariant subspaces are direct sums of eigenspaces, and any basis for this space can be extended to an eigenbasis.
38.
Lie's theorem states that any nonzero representation of a solvable Lie algebra on a finite dimensional vector space over an algebraically closed field of characteristic 0 has a one-dimensional invariant subspace.
39.
It follows that any closed invariant subspace is generated by the algebraic direct sum of eigenspaces it contains and that this sum is invariant under the infinitesimal action of the Lie algebra \ mathfrak g.
40.
Each of these has a center that is easy to describe . ( 2 ) For an algebra that is not semisimple, there are non-trivial invariant subspaces by Nakayama's lemma.
