Hilbert Space Operators

Let χ be an inner product space. Recall that |〈x;y〉| ≤ ||x|| ||y|| (Schwartz inequality) and ||x+y||2 = ||x||2+2 Re 〈 x ; y 〉 + ||y||2 for every x and y in χ, where the norm || || is that induced by the inner product 〈;〉. Two vectors x and y in χ are orthogonal if 〈 x; y 〉 = 0. In this case we write x ⊥ y. Two subsets A and В of χ are orthogonal (notation: A ⊥ В) if every vector in A is orthogonal to every vector in В. The orthogonal complement of a set A is the set A⊥ made up of all vectors in χ that are orthogonal to every vector of A. Observe that {0}⊥= χ, χ⊥ = {0}, and A ⊥ В if and only if $$A \subseteq {B^ \bot }$$ . Moreover, A⊥ is a subspace (closed linear manifold) of χ, and $$A \cap {A^ \bot } \subseteq \{ 0\}$$ {0}. In fact, A⊥ = (А⊥)- = (А-)⊥. If М is a linear manifold of χ,then М ⋂ М⊥= {0} and, if χ is a Hilbert space, then М⊥⊥ = М-, and М⊥ = {0} if and only if, М- = χ (see e.g., [32, §5.4]).