Spectral Theory in Hilbert Space

96 13. FIRST ORDER SYSTEMS
Clearly all these restrictions are obtained by restricting the boundary
values of elements in T1 to certain n-dimensional subspaces of C 2n ,
i.e., by imposing n linear, homogeneous boundary conditions on T1.
We consider a few special cases. One selfadjoint realization is obtained
by imposing periodic boundary conditions u(b) = u(a) or more generally
u(b) = Su(a) where S is a fixed matrix satisfying S ∗ JS = J. As
already mentioned, such a matrix S is often called symplectic, at least
, so that n is even.
in the case when S is real, and J = 0 I
−I 0
Another possibility occurs if the invertible Hermitian matrix iJ has
an equal number of positive and negative eigen-values (this obviously
requires n to be even). In that case we may impose separated boundary
conditions, i.e., conditions that make both u∗ (a)Ju(a) and u∗ (b)Ju(b)
vanish. Boundary conditions which are not separated are called coupled.
It must be emphasized that for n > 2 there are selfadjoint realizations
which are determined by some conditions imposed only on the
value at one of the endpoints, and some conditions involving the values
at both endpoints.
Let us now turn to the general, not necessarily regular case. We
first need to briefly discuss Hermitian forms of finite rank. If B is a
Hermitian form on a linear space L we set LB = {u ∈ L | B(u, L) = 0}
which is a subspace of L. The rank of B is codim LB (= dim L/LB ).
In the sequel we assume that B has finite rank. If M is a subspace on
which the form B is non-degenerate, i.e., there is no non-zero element
u ∈ M such that B(u, v) = 0 for all v ∈ M, then we must have
LB ∩ M = {0} so that M has to be finitedimensional. This means, of
course, that after introducing a basis in M the form B is given on M
by an invertible matrix. If B is non-degenerate on M, then for every
u ∈ L there is a unique element v ∈ M (the B-projection of u on M)
such that B(u − v, M) = 0 (Exercise 13.5). If B is non-degenerate on
M, but not on any proper superspace of M, we say that M is maximal
non-degenerate for B. Of course this means exactly that LB ∩M = {0}
and dim M = rank B so that L = M ˙+L B as a direct sum.
We call a subspace P of L on which B is positive definite a maximal
positive definite space for B if P has no proper superspaces on
which B is positive definite. If B is positive definite on P, then clearly
dim P ≤ rank B. It follows that forms of finite rank always have maximal
positive definite spaces. Similarly for negative definite spaces.
Proposition 13.10 (Sylvester’s law of inertia). Suppose B is a
Hermitian form of finite rank on a linear space L. Then all maximal
positive definite subspaces for B have the same dimension. Similarly
for maximal negative definite subspaces.
Proof. Suppose P is maximal positive definite for B and that ˜ P
is another positive definite space for B. Then the B-projection on P
is injective as a linear map BP : ˜ P → P. For if not, there exists