Page 262 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 24: Orthogonal Complements

Two non-empty subsets S and S of a Hilbert space H are said to be orthogonal denoted by Notes
1 2
S S , if x y for every x S and every y S .
1 2 1 2
24.1.4 Orthogonal Compliment: Definition

Let S be a non-empty subset of a Hilbert space H. The orthogonal compliment of S, written as S
and is read as S perpendicular, is defined as
S = {x H : x y y S}

Thus, S is the set of all those vectors in H which are orthogonal to every vectors in H which are
orthogonal to every vector in S.
Theorem 1: If S, S , S are non-empty subsets of a Hilbert space H, then prove the following:
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(a) {0} = H (b) H = {0} (c) S S {0}
(d) S S S S (e) S S
1 2 2 1
Proof:
(a) Since the orthogonal complement is only a subset of H, {0} H.
It remains to show that H {0} .

Let x H. Since (x, 0) = 0, therefore x {0} .
Thus x H x {0} .
H {0} .
Hence {0} = H
(b) Let x H. Then by definition of H, we have

(x, y) = 0 y H
Taking y = x, we get

2
(x, x) = 0 x = 0 x = 0
Thus x H x = 0
H = {0}
(c) x S S .

Then x S and x S
Since x S , therefore x is orthogonal to every vector in S. In particular, x is orthogonal to
x because x S.
2
Now (x, x) = 0 x = 0 x = 0.
0 is the only vector which can belong to both S and S .
S S {0}
If S is a subspace of H, then 0 S. Also S is a subspace of H. Therefore 0 S . Thus, if S is
a subspace of H, then 0 S S . Therefore, in this case S S = {0}.

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