## 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.

### Invariant

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.

### Complementary

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.