Spectral Theory in Hilbert Space

6 1. LINEAR SPACES
of u so one may also regard C n as the set of ordered n-tuples of complex
numbers. The corresponding real space is the usual R n .
If L is a linear space and V a subset of L which is itself a linear
space, using the linear operations inherited from L, one says that V is
a linear subspace of L.
Proposition 1.2. A non-empty subset V of L is a linear subspace
of L if and only if u + v ∈ V and λu ∈ V for all u, v ∈ V and λ ∈ C.
The proof is left as an exercise (Exercise 1.3). If u1, u2, . . . , uk are
elements of a linear space L we denote by [u1, u2, . . . , uk] the linear
hull of u1, u2, . . . , uk, i.e., the set of all linear combinations λ1u1 +· · ·+
λkuk, where λ1, . . . , λk ∈ C. It is not hard to see that linear hulls are
always subspaces (Exercise 1.5). One says that u1, . . . , uk generates
L if L = [u1, . . . , uk], and any linear space which is the linear hull
of a finite number of its elements is called finitely generated or finitedimensional.
A linear space which is not finitely generated is called
infinite-dimensional. It is clear that if, for example, u1 is a linear
combination of u2, . . . , uk, then [u1, . . . , uk] = [u2, . . . , uk]. If none of
u1, . . . , uk is a linear combination of the others one says that u1, . . . , uk
are linearly independent. It is clear that any finitely generated space
has a set of linearly independent generators; one simply starts with
a set of generators and goes through them one by one, at each step
discarding any generator which is a linear combination of those coming
before it. A set of linearly independent generators for L is called a basis
for L. A given finite-dimensional space L can of course be generated
by many different bases. However, a fundamental fact is that all such
bases of L have the same number of elements, called the dimension of
L. This follows immediately from the following theorem.
Theorem 1.3. Suppose u1, . . . , uk generate L, and that v1, . . . , vj
are linearly independent elements of L. Then j ≤ k.
Proof. Since u1, . . . , uk generate L we have v1 = k
s=1 x1sus,
for some coefficients x11, . . . , x1k which are not all 0 since v1 = 0.
By renumbering u1, . . . , uk we may assume x11 = 0. Then u1 =
1
x11 v1 − k x1s
s=2 x11 us, and therefore v1, u2, . . . , uk generate L. In particular,
v2 = x21v1 + k s=2 x2sus for some coefficients x21, . . . , x2k. We
can not have x22 = · · · = x2k = 0 since v1, v2 are linearly independent.
By renumbering u2, . . . , uk, if necessary, we may assume x22 = 0. It
follows as before that v1, v2, u3, . . . , uk generate L. We can continue in
this way until we run out of either v:s (if j ≤ k) or u:s (if j > k). But
if j > k we would get that v1, . . . , vk generate L, in particular that
vj is a linear combination of v1, . . . , vk which contradicts the linear
independence of the v:s. Hence j ≤ k.
For a finite-dimensional space the existence and uniqueness of coordinates
for any vector with respect to an arbitrary basis now follows