Jordan Forms, Direct Sum, Invariant, Complementary

Direct Sum of Vector Spaces

The span of two subspaces U and V is the smallest subspace containing both. This is the set of vectors x+y such that x ∈ U and y ∈ V. Verify that this is indeed a subspace, and it must be included in any subspace containing U and V, hence it is the span of U and V.

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.


A subspace is invariant under a transformation if that subspace is mapped into itself. I don't particularly like this definition, because it sounds like the subspace is fixed by the transformation, and that isn't necessarily the case. If a transformation rotates space about the z axis, The z axis is fixed, and the xy plane is invariant. So the word invariant is a little confusing, but to be fair, I can't think of another adjective that would be any better, so invariant it is.

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.


Subspaces are complementary if they are invariant and their direct sum produces the original space. Note that spaces are complementary with respect to a given transformation.

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.