Final answer:
The intersection of any collection of T-invariant subspaces of V is T-invariant because the intersection is closed under vector addition and scalar multiplication, and applying T to any vector in the intersection yields a vector still in the intersection.
Step-by-step explanation:
To prove that the intersection of any collection of T-invariant subspaces of V is a T-invariant subspace of V, we need to verify two things. First, that the intersection is indeed a subspace, and second, that it is T-invariant. A subspace must be closed under vector addition and scalar multiplication. If we take any two vectors in the intersection, they must be in all T-invariant subspaces by definition of intersection. Since each T-invariant subspace is closed under vector addition and scalar multiplication, our vectors' sum and scalar multiples will also be in each T-invariant subspace, and hence in the intersection. Therefore, the intersection is a subspace.
Now, to show T-invariance, let's take any vector in the intersection of T-invariant subspaces and apply T to it. Since the vector is in each T-invariant subspace, T applied to the vector must produce a vector that is still within each T-invariant subspace. Thus, T(v) is in the intersection. This proves that the intersection of any collection of T-invariant subspaces is itself a T-invariant subspace.