If U and V are disjoint (except for 0), then the span is called the sum, or the direct sum, or the direct product. Every vector in the span has a unique representation as x+y, where x comes from U and y comes from V. If this fails then x1+y1 = x2+y2, hence x1-x2 = y2-y1, and U and V intersect after all. We can also claim U and V are linearly independent. If x+y = 0 then x = -y, U contains -y, U contains y, and U and V intersect. To complete the circle, assume U and V are linearly independent. If they are not disjoint, if they share a common vector x, then we could write x + -x = 0, which contradicts independence.
In summary, the span is a direct sum (i.e. every vector in the span has a unique representation), iff U and V are disjoint, iff U and V are linearly independent spaces.
some people write the direct sum using the plus operator, as in U+V. Others write it as U*V. I prefer the * notation, because the span really is a cross product of all possible vectors in both sets. Granted, the vectors are added together, as part of an abelian group, so + makes sense too; but it's still a cross product, and I lean towards U*V.
If U and V are independent, a basis for U and a basis for V combine to form a basis for U*V.
Given a linear transformation and an eigen value w, the eigen vectors with eigen value w form an invariant vector space. In fact each vector in that space remains in position; it is simply scaled by the eigen value w.
Zero is always invariant.
Show that the span and intersection of two invariant subspaces is an invariant subspace.
If our basis consists of a basis for U and a basis for V, and U and V are complementary, the transformation is implemented by a block diagonal matrix, where the upper left block operates on U, and the lower right block operates on V.