Orthogonal Complements and Projections
1. Definition of the Orthogonal Complement
Given a subspace \(M\) of a Hilbert space \(H\), we define its orthogonal complement \(M^\perp\) to be the collection of all those vectors \(y \in H\) such that \(\left<x,y\right> = 0\) for all \(x \in M\).
Theorem
The orthogonal complement \(M^\perp\) of a closed subspace \(M\) satisfies:
- \(M^\perp\) is a closed subspace.
- \(M^\perp \cap M = \{0\}\).
- \((M^\perp)^{\perp} = M\).
- \(M + M^{\perp} = H\).
Proof
2. A Lemma: Minimizing Distance to Closed, Convex Sets
Lemma
Suppose \(E\) is a closed convex set in a Hilbert space \(H\) and \(v\) is a vector in \(H\) not belonging to \(E\). Then there exists a unique \(e_* \in E\) such that \(||e_* - v|| = \inf_{e \in E} ||e - v||\) (i.e., the infimum is attained).
Proof
3. Orthogonal Projections
Theorem (Projections)
Let \(M\) be a closed subspace of a Hilbert space \(H\). Then the following are true:
- Every \(x \in H\) has a unique decomposition \(x = P x + Q x\) where \(Px \in M\) and \(Q x\) is orthogonal to every \(y \in M\).
- The mappings \(x \mapsto P x\) and \(x \mapsto Q x\) are both linear.
- \(||x||^2 = ||Px||^2 + ||Qx||^2\).
- \(P^2 x = P x\), \(Q^2 x = Q x\), \(P Q x = Q P x = 0\).
Proof
Proposition
Suppose that \(M_1\) and \(M_2\) are closed subspaces. Then
- \(M_1 \cap M_2\) is closed.
- \((\overline{M_1 + M_2})^\perp = M_1^\perp \cap M_2^\perp\)
- \((M_1 \cap M_2)^\perp = \overline{M_1^\perp + M_2^\perp}\)
Proof
Important (Sums of closed subspaces aren't always closed)
Let \(H := \ell^2({\mathbb N})\). Let \(M\) be the closed subspace of all sequences
\[ (a_1,0,a_3,0,a_5,0,\ldots) \in \ell^2({\mathbb N}). \]
Let \(N\) be the closed subspace of all sequences
\[ \left(b_1,\frac{b_1}{2},b_3,\frac{b_3}{4},b_5,\frac{b_5}{6},\ldots \right) \in \ell^2({\mathbb N}). \]
The intersection \(M \cap N\) is trivial, so every vector in \(M + N\) has a unique decomposition. For the vector
\[{}v_n{}\]
\[{}:={}\]
\[{}\Big(0,1,0,\frac{1}{2},0,\frac{1}{4},{}\]
\[{}\ldots,0,\frac{1}{2n},0,0,0,\ldots \Big){}\]
it is
\[{}(-1,0,-1,0, \ldots,-1,0,0,0,\ldots){}\]
\[{} + \left(1,1,1,\frac{1}{2},1,\frac{1}{4},\ldots,1,\frac{1}{2n},0,\ldots \right).{}\]
But \(\lim_{n \rightarrow \infty} v_n\) exists but the limit vector \(v\) does not belong to the sum \(M + N\) because every sequence \((c_1,c_2,\ldots)\) in \(M + N\) necessarily has \(2n c_{2n} - c_{2n-1} \rightarrow 0\) as \(n \rightarrow \infty\).