Orthogonal complement
Given an inner product space
Proof of subspace
Clearly\Span
. If and then for all , and thus . Therefore is a subspace.
Properties
Let
is topologically closed - If
for some , then
Proof of 1–6
Note that the orthogonal complement of a singleton
can be expressed as a preïmage Since
is closed in , and the inner product is continuous, it follows is closed. Now for an arbitrary set , which is an intersection of closed sets and is therefore closed, proving ^S1.
Note if
then , implying by ^IP3, proving ^S2. Let
and . Then for any , so , proving ^S3. Let
. Then by definition for any , so , proving ^S4. Without loss of generality
, for iff . Now and
but since
, it follows from ^IP3 that , proving ^S5. Let
and , so for some . Then proving ^S6.
Let
(Internal direct sum). .
Proof of 1–2
Other properties include