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单词 ComplementarySubspace
释义

complementary subspace


Direct sum decomposition.

Let U be a vector spaceMathworldPlanetmath, and V,WU subspacesPlanetmathPlanetmath. We say thatV and W span U, and write

U=V+W

ifevery uU can be expressed as a sum

u=v+w

for some vV and wW.

If in additionPlanetmathPlanetmath, such a decomposition is unique for all uU, orequivalently if

VW={0},

then we say that V and W form a direct sumMathworldPlanetmath decomposition of Uand write

U=VW.

In such circumstances, we also say that V and Ware complementary subspaces, and also say that W is an algebraic complement of V.

Here is useful characterization of complementary subspaces if U isfinite-dimensional.

Proposition 1

Let U,V,W be as above, and suppose that U isfinite-dimensional. The subspaces V and W are complementary ifand only if for every basis v1,,vm of V andevery basisw1,,wn of W, the combined list

v1,,vm,w1,,wn

is a basis of U.

Remarks.

  • Since every linearly independent subset of a vector space can be extended to a basis, every subspace has a complementPlanetmathPlanetmath, and the complement is necessarily unique.

  • Also, direct sum decompositions of a vector space U are in a one-to correspondence fashion with projections on U.

Orthogonal decomposition.

Specializing somewhat, suppose that the ground field 𝕂 is eitherthe real or complex numbers, and that U is either an inner productspaceMathworldPlanetmath or a unitary space, i.e. U comes equipped with apositive-definite inner productMathworldPlanetmath

,:U×U𝕂.

In such circumstances,for every subspace VU we define the orthogonal complementMathworldPlanetmath ofV, denoted by V to be the subspace

V={uU:v,u=0, for all vV}.
Proposition 2

Suppose that U is finite-dimensional and VU a subspace.Then, V and its orthogonalcomplement V determine a direct sum decomposition of U.

Note: the PropositionPlanetmathPlanetmath is false if either the finite-dimensionalityor the positive-definiteness assumptionsPlanetmathPlanetmath are violated.

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更新时间:2025/5/4 2:36:52