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In the mathematical fields of linear algebra and functional analysis, the orthogonal complement of a subspace of a vector space equipped with a bilinear form is the set of all vectors in that are orthogonal to every vector in .
In a three-dimensional Euclidean vector space, the orthogonal complement of a line through the origin is the plane through the origin perpendicular to it, and vice versa. [ 5 ] Note that the geometric concept of two planes being perpendicular does not correspond to the orthogonal complement, since in three dimensions a pair of vectors, one from ...
In a Hilbert space, the orthogonal complement of any closed vector subspace is always a topological complement of . This property characterizes Hilbert spaces within the class of Banach spaces : every infinite dimensional, non-Hilbert Banach space contains a closed uncomplemented subspace, a deep theorem of Joram Lindenstrauss and Lior Tzafriri .
2. Orthogonal subspace in the dual space: If W is a linear subspace (or a submodule) of a vector space (or of a module) V, then may denote the orthogonal subspace of W, that is, the set of all linear forms that map W to zero. 3. For inline uses of the symbol, see ⊥.
The Schur complement is named after Issai Schur [1] who used it to prove Schur's lemma, although it had been used previously. [2] Emilie Virginia Haynsworth was the first to call it the Schur complement. [3] The Schur complement is sometimes referred to as the Feshbach map after a physicist Herman Feshbach. [4]
Several sets of orthogonal functions have become standard bases for approximating functions. For example, the sine functions sin nx and sin mx are orthogonal on the interval x ∈ ( − π , π ) {\displaystyle x\in (-\pi ,\pi )} when m ≠ n {\displaystyle m\neq n} and n and m are positive integers.
Two vectors and are orthogonal if , =, which happens if and only if ‖ ‖ ‖ + ‖ for all scalars . [2] The orthogonal complement of a subset is := {: , =}, which is always a closed vector subspace of .
If a normal operator T on a finite-dimensional real [clarification needed] or complex Hilbert space (inner product space) H stabilizes a subspace V, then it also stabilizes its orthogonal complement V ⊥. (This statement is trivial in the case where T is self-adjoint.) Proof. Let P V be the orthogonal projection onto V.