## Linear Algebra: Invariant Subspaces

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. Assume that V is finite dimensional.

The attempt at a solution

I really think that I should be able to produce a counterexample, however even if I find an appropriate subspace, I have NO idea how to show that U is invariant under every operator on V. There are an arbitrary number of them!

One thing I was thinking was use a one dimensional subspace for my U, because then every transformation of a vector u in U would be a constant times u. In other words, T(u) = cu.

Does anyone know if I'm on the right track?
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 Recognitions: Homework Help Science Advisor Hint #1: You're not going to be able to produce a counterexample. Hint #2: Suppose U is a nontrivial subspace of V that's invariant under every operator on V, and let {u_1, ..., u_n} be a basis for U. A linear operator is completely determined by its action on the basis.
 My textbook states that for operators on complex vector spaces with dimension greater than one, and real vector spaces with dimension greater than two, that there will be invariant subspaces other than {0} and V. Maybe the book means for a particular operator?

Recognitions:
Homework Help
 Recognitions: Gold Member Science Advisor Staff Emeritus morphism's point: Given [any subspace U of V, of dimension m, 0< m< n= dim V, there exist a basis for V, $v_1, v_2, ..., v_n}$ such that the first m vectors, [itex]{v_1, v_2, ..., v_m} form a subspace for U. Now CONSTRUCT a linear transformation on V such that U is NOT invariant under that linear transformation.