Additive map

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In algebra, an additive map, -linear map or additive function is a function f that preserves the addition operation:[1] f(x+y)=f(x)+f(y) for every pair of elements x and y in the domain of f. For example, any linear map is additive. When the domain is the real numbers, this is Cauchy's functional equation. For a specific case of this definition, see additive polynomial.

More formally, an additive map is a -module homomorphism. Since an abelian group is a -module, it may be defined as a group homomorphism between abelian groups.

A map V×WX that is additive in each of two arguments separately is called a bi-additive map or a -bilinear map.[2]

Examples

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Typical examples include maps between rings, vector spaces, or modules that preserve the additive group. An additive map does not necessarily preserve any other structure of the object; for example, the product operation of a ring.

If f and g are additive maps, then the map f+g (defined pointwise) is additive.

Properties

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Definition of scalar multiplication by an integer

Suppose that X is an additive group with identity element 0 and that the inverse of xX is denoted by x. For any xX and integer n, let: nx:={0 when n=0x++x(n summands)  when n>0(x)++(x)(|n| summands)  when n<0. Thus (1)x=x and it can be shown that for all integers m,n and all xX, (m+n)x=mx+nx and (nx)=(n)x=n(x). This definition of scalar multiplication makes the cyclic subgroup x of X into a left -module; if X is commutative, then it also makes X into a left -module.

Homogeneity over the integers

If f:XY is an additive map between additive groups then f(0)=0 and for all xX, f(x)=f(x) (where negation denotes the additive inverse) and[proof 1] f(nx)=nf(x) for all n. Consequently, f(xy)=f(x)f(y) for all x,yX (where, by definition, xy:=x+(y)).

In other words, every additive map is homogeneous over the integers. Consequently, every additive map between abelian groups is a homomorphism of -modules.

Homomorphism of -modules

If the additive abelian groups X and Y are also a unital modules over the rationals (such as real or complex vector spaces) then an additive map f:XY satisfies:[proof 2] f(qx)=qf(x) for all q and xX. In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital -modules is a homomorphism of -modules.

Despite being homogeneous over , as described in the article on Cauchy's functional equation, even when X=Y=, it is nevertheless still possible for the additive function f: to not be homogeneous over the real numbers; said differently, there exist additive maps f: that are not of the form f(x)=s0x for some constant s0. In particular, there exist additive maps that are not linear maps with respect to an existing ring structure of the codomain.

See also

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Notes

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  1. ^ Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).
  2. ^ Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).

Proofs

  1. ^ f(0)=f(0+0)=f(0)+f(0) so adding f(0) to both sides proves that f(0)=0. If xX then 0=f(0)=f(x+(x))=f(x)+f(x) so that f(x)=f(x) where, by definition, (1)f(x):=f(x). Induction shows that if n is positive then f(nx)=nf(x) and that the additive inverse of nf(x) is n(f(x)), which implies that f((n)x)=f(n(x))=nf(x)=n(f(x))=(nf(x))=(n)f(x) (this shows that f(nx)=nf(x) holds for n<0).
  2. ^ Let xX and q=mn where m,n and n>0. Let y:=1nx. Then ny=n(1nx)=(n1n)x=(1)x=x, which implies f(x)=f(ny)=nf(y)=nf(1nx) so that multiplying both sides by 1n proves that f(1nx)=1nf(x). Consequently, f(qx)=f(mnx)=mf(1nx)=m(1nf(x))=qf(x).

References

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  • Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).