# Invariant subspaces under linear operators

1. Sep 30, 2008

### jimmypoopins

1. The problem statement, all variables and given/known data
Prove or give a counterexample: If U is a subspace of V that is invariant under every operator on V, then U = {0} or U = V.

2. Relevant equations
U is invariant under a linear operator T if u in U implies T(u) is in U.

3. The attempt at a solution
Assume {0} does not equal U does not equal V. Let {u1,...,un} be a basis for U. Extend to a basis for V: {u1,...,un,v1,...,vm}. Since V does not equal {0}, m is greater than or equal to 1. Define a linear operator by T=v1, i=1,...n and T(vi)=v1, i=1,...,m. Then U is not invariant under T.

I think this is a counterexample to the contrapositive of the statement. does it work? (the contrapositive is If U does not equal {0} does not equal V, then U is not invariant under every operator on V, right?)

2. Oct 1, 2008

### morphism

I think you have the right idea, but your write up has some typos in it so I'm not sure.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook