Invariance - Normal Linear Transformation

[e(ho0n3]

Homework Statement
Let W be a complex finite dimensional vector space with a hermitian scalar product and let T: W -> W be linear and normal. Prove that U is a T-invariant subspace of W if and only if V is a T*-invariant subspace, where V is the orthogonal complement of U.

The attempt at a solution
Let u in U and v in V. If U is T-invariant, then (Tu, v) = 0. But (Tu, v) = (u, T*v) so T*v is in V and V is T*-invariant. Similarly, if V is T*-invariant, then U is T-invariant. Note that I haven't used the fact that T is normal. It seems to me that T doesn't have to be normal. Am I wrong?

 

[morphism]

No, you are absolutely correct. However I think there is a typo in the problem: it should have said "V is a T-invariant subspace" instead of "V is a T*-invariant subspace" (which is trivial, as you noted). This modified exercise is a bit more challenging, and does require the fact that T is normal. :)

 

[e(ho0n3]

How did you spot that? I'll try to solve the modified exercise.

 

[morphism]

I spotted it because I'm an invariant subspace aficionado. ;) In fact the converse is also true: if an operator T on a finite dimensional complex inner product space has the property that $U^\perp$ is T-invariant whenever U is, then T must be normal.

A subspace U is said to be reducing if both it and its ortho complement are invariant. The reason for this terminology follows from the observation that if U is reducing for T, then the matrix for T with respect to the decomposition $U \oplus U^\perp$ is block diagonal, so that in some sense U "reduces" T to something simpler. We can thus restate your exercise and its converse succinctly as: T is normal iff every T-invariant subspace is reducing.

...at least this is the case in finite dimensions. In infinite dimensions, i.e. in the setting of Hilbert spaces, it is actually not known if this is true; in fact, this problem is equivalent to one of the most notorious open problems in analysis: the invariant subspace problem.