# Complementary space

A complementary subspace short complementary space or complement is, in mathematical branch of linear algebra as large as possible subspace of the vector space that intersects a predetermined sub-space only at the origin. The entire vector space is thus divided into two independent parts.

## Complement of a subspace

### definition

Let it be a vector space over a field and a subspace of . Then a subspace is called complementary or a complement to if the conditions ${\ displaystyle V}$ ${\ displaystyle K}$${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle W}$${\ displaystyle U}$

• ${\ displaystyle U \ cap W = \ {0 \}}$

and

• ${\ displaystyle U + W = V}$

are fulfilled. It is the zero vector space and is short for ${\ displaystyle \ {0 \}}$${\ displaystyle U + W}$

${\ displaystyle \ {u + w \ mid u \ in U, w \ in W \}.}$

• One then also says: is the inner direct sum of and and writes .${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle W}$${\ displaystyle V = U \ oplus W}$
• If subspaces of and are their outer direct sum, then: The homomorphism${\ displaystyle U, W}$${\ displaystyle V}$${\ displaystyle U \ oplus W}$
${\ displaystyle U \ oplus W \ to V, \ \ (u, w) \ mapsto u + w}$
is an isomorphism if and only if and are complementary, d. H. if is the inner direct sum of and .${\ displaystyle U}$${\ displaystyle W}$${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle W}$
• A complementary subspace always exists for a subspace of a vector space. This follows from the basic supplementary sentence . However, complements are generally not clearly defined.${\ displaystyle U}$${\ displaystyle V}$
• ${\ displaystyle W}$is a complement of in if and only if each vector is uniquely${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle v \ in V}$
${\ displaystyle v = u + w}$
with and lets write.${\ displaystyle u \ in U}$${\ displaystyle w \ in W}$
• The following applies for the dimensions of the corresponding subspaces
${\ displaystyle \ dim V = \ dim U + \ dim W.}$
The dimension of the complementary space is also referred to as the codimension of in .${\ displaystyle W}$${\ displaystyle U}$${\ displaystyle V}$
• If a complement is closed, a complement is also closed .${\ displaystyle W}$${\ displaystyle U}$${\ displaystyle U}$${\ displaystyle W}$
• The restriction of the canonical projection to is an isomorphism , see factor space .${\ displaystyle V \ to V / U}$${\ displaystyle W}$

### Connection with projections

Let it be a subspace in vector space . ${\ displaystyle U}$${\ displaystyle V}$

• If is a complementary space of , then, according to the above, each element of can be represented uniquely as a sum with and . Then there is a projection with the picture and core .${\ displaystyle W}$${\ displaystyle U}$${\ displaystyle v}$${\ displaystyle V}$${\ displaystyle v = u + w}$${\ displaystyle u \ in U}$${\ displaystyle w \ in W}$${\ displaystyle P_ {W} \ colon V \ rightarrow V, v = u + w \ mapsto u}$ ${\ displaystyle \ operatorname {im} P_ {W} = U}$ ${\ displaystyle \ operatorname {ker} P_ {W} = W}$
• Conversely , if there is a projection with an image , the core is a complementary space of .${\ displaystyle P \ colon V \ rightarrow V}$${\ displaystyle U}$${\ displaystyle \ operatorname {ker} P}$${\ displaystyle U}$

In this way a bijection is obtained from the set of all complementary spaces from onto the set of all projections onto with image . The projections with image form an affine space above the vector space . ${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle U}$${\ displaystyle \ operatorname {Hom} (V / U, U) \ subset \ operatorname {Hom} (V, V)}$

Every subspace is a complement to it .${\ displaystyle W_ {a}}$${\ displaystyle U}$

### example

We consider the subspace as in the adjacent drawing. For every real number, let the straight line through 0 with a slope . Any such subspace is too complementary a subspace of . The associated projection has the matrix representation . You can see directly from the matrix display that the picture is, because the first line of the matrix consists only of zeros. The kernel of is , because from follows , that is, the kernel consists of all points with , and that is exactly the straight line through 0 with a slope . ${\ displaystyle U: = \ {(0, y); \, y \ in \ mathbb {R} \} \ subset V = \ mathbb {R} ^ {2}}$${\ displaystyle a}$${\ displaystyle W_ {a}}$${\ displaystyle a}$${\ displaystyle W_ {a}}$${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle P_ {a} = {\ begin {pmatrix} 0 & 0 \\ - a & 1 \ end {pmatrix}}}$${\ displaystyle U}$${\ displaystyle P_ {a}}$${\ displaystyle W_ {a}}$${\ displaystyle P_ {a} {\ begin {pmatrix} x \\ y \ end {pmatrix}} = {\ begin {pmatrix} 0 \\ 0 \ end {pmatrix}}}$${\ displaystyle {\ begin {pmatrix} 0 \\ 0 \ end {pmatrix}} = {\ begin {pmatrix} 0 & 0 \\ - a & 1 \ end {pmatrix}} {\ begin {pmatrix} x \\ y \ end { pmatrix}} = {\ begin {pmatrix} 0 \\ - ax + y \ end {pmatrix}}}$${\ displaystyle {\ begin {pmatrix} x \\ y \ end {pmatrix}}}$${\ displaystyle y = ax}$${\ displaystyle a}$

## Orthogonal complement

### definition

Let it be a vector space over a body on which a symmetrical or alternating bilinear form or a Hermitian sesquilinear form is given. For a subspace means ${\ displaystyle V}$${\ displaystyle K}$ ${\ displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle U \ subseteq V}$

${\ displaystyle U ^ {\ perp}: = \ {v \ in V \ mid \ forall u \ in U: \ langle u, v \ rangle = 0 \}}$

the orthogonal complement or the orthogonal space of in . Note that it is generally not a complement of as defined above. The duality theorem says, however, that if it is finite-dimensional and has not degenerated both on and on the subspace , then applies. ${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle V}$${\ displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle V}$${\ displaystyle U}$ ${\ displaystyle V = U \ oplus U ^ {\ bot}}$

The last property is always fulfilled , for example, for scalar products on real or complex vector spaces.

### Orthogonal complement in Hilbert spaces

If a Hilbert space is , then the orthogonal complement of a subspace is a complement of its closure , i. H. ${\ displaystyle V}$${\ displaystyle U}$ ${\ displaystyle {\ bar {U}}}$

${\ displaystyle V = {\ bar {U}} \ oplus U ^ {\ perp}}$, where can be read as an inner orthogonal sum .${\ displaystyle \ oplus}$

The orthogonal complement is always closed, and it holds

${\ displaystyle (U ^ {\ perp}) ^ {\ perp} = {\ bar {U}}}$.

## Complements in Banach spaces

Let a (finite-dimensional or infinite-dimensional) complete, normalized vector space, i.e. a Banach space , be a closed subspace to which a closed complementary space exists, so that the spaces and are algebraically isomorphic, then the isomorphism defined by is also a topological isomorphism. That is, the mapping and its inverse mapping are continuous . ${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle W}$${\ displaystyle V}$${\ displaystyle U \ oplus W}$${\ displaystyle U \ oplus W \ rightarrow V, \, (u, w) \ mapsto u + w}$

In Banach spaces, closed subspaces according to the above always have a complementary space, but that does not mean that one could also find a closed complementary space. Rather, this is a characterization of the topological vector space structure of Hilbert spaces , in which the orthogonal complement is always available, because the following theorem of Lindenstrauss and Tzafriri applies :

• A Banach space is continuously isomorphic to a Hilbert space if and only if every closed subspace has a closed complementary space.

The following Sobczyk theorem applies to the existence of complementary spaces :

In the not-necessarily-separable case, however, the statement does not apply: One can show that there is no closed complementary space for. ${\ displaystyle c_ {0} \ subset \ ell ^ {\ infty}}$

## Invariant complements

Let be a vector space, an endomorphism of and an -invariant subspace, i. H. . Then it does n't always have an -invariant complement. If there is an invariant complement for every invariant subspace, the endomorphism is called semi- simple . Over algebraically closed fields , semi-simplicity is equivalent to diagonalizability . ${\ displaystyle V}$${\ displaystyle f \ colon V \ to V}$${\ displaystyle V}$${\ displaystyle U}$${\ displaystyle f}$${\ displaystyle f (U) \ subseteq U}$${\ displaystyle U}$${\ displaystyle f}$

Analogous terms are used in representation theory. For a unitary representation , the orthogonal complement of an invariant subspace is again invariant, consequently every finite-dimensional unitary representation is semi-simple.

If the invariant subspaces are interpreted as sub-modules, the invariant complements become complementary sub-modules in the sense of the following section.

## generalization

The definition of complements can literally be generalized to modules . However, there is no longer always a complementary sub-module to a sub-module of a module over a ring. A module in which each sub-module has a complement is called a semi-simple module . In this way of speaking, for example, vector spaces are semi-simple modules. The module is not semi-simple because the sub-module has no complement. ${\ displaystyle \ mathbb {Z}}$${\ displaystyle \ mathbb {Z}}$${\ displaystyle 2 \ mathbb {Z}}$

Instead of “has a complement” one also says “is a direct summand”. Projective modules are characterized by the fact that they are isomorphic to direct summands of free modules. Injective modules are characterized by the fact that they have a complement in each main module.

The relationship to projections as well as the simply transitive operation of on the set of complements of in is also transferred to the module case (even to any Abelian categories ). ${\ displaystyle \ operatorname {Hom} (V / U, U)}$${\ displaystyle U}$${\ displaystyle V}$