An important fact about the category of vector spaces is that all exact sequences split. That is, if we have a short exact sequence
we can find a linear map from 
to 
which lets us view it as a subspace of , and we can write 
. When we have an inner product around and is finite-dimensional, we can do this canonically.
What we’ll do is define the orthogonal complement of 
to be the vector space
That is, 
consists of all vectors in perpendicular to every vector in 
.
First, we should check that this is indeed a subspace. If we have vectors 
, scalars 
, and a vector 
, then we can check
and thus the linear combination 
is also in .
Now to see that 
, take an orthonormal basis 
for . Then we can expand it to an orthonormal basis 
of . But now I say that 
is a basis for . Clearly they’re linearly independent, so we just have to verify that their span is exactly .
First, we can check that 
for any 
between 
and 
, and so their span is contained in . Indeed, if 
is a vector in , then we can calculate the inner product
since 
and 
. Of course, we omit the conjugation when working over 
.
Now, let’s say we have a vector 
. We can write it in terms of the full basis 
as 
. Then we can calculate its inner product with each of the basis vectors of as
Since this must be zero, we find that the coefficient 
of 
must be zero for all 
between 
and 
. That is, is contained within the span of
So between a basis for and a basis for we have a basis for with no overlap, we can write any vector 
uniquely as the sum of one vector from and one from , and so we have a direct sum decomposition as desired.
May 4, 2009
Posted by John Armstrong |
Algebra, Linear Algebra |
12 Comments
