Additive_map Knowpia

In algebra, an additive map, $Z$ -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 $\mathbb {Z}$ -module homomorphism. Since an abelian group is a $\mathbb {Z}$ -module, it may be defined as a group homomorphism between abelian groups.

A map $V\times W\to X$ that is additive in each of two arguments separately is called a bi-additive map or a $\mathbb {Z}$ -bilinear map.^[2]

Examples edit

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 edit

Definition of scalar multiplication by an integer

Suppose that $X$ is an additive group with identity element $0$ and that the inverse of $x\in X$ is denoted by $-x.$ For any $x\in X$ and integer $n\in \mathbb {Z} ,$ let:

nx:=\left\{{\begin{alignedat}{9}&&&0&&&&&&~~~~&&&&~{\text{ when }}n=0,\\&&&x&&+\cdots +&&x&&~~~~{\text{(}}n&&{\text{ summands) }}&&~{\text{ when }}n>0,\\&(-&&x)&&+\cdots +(-&&x)&&~~~~{\text{(}}|n|&&{\text{ summands) }}&&~{\text{ when }}n<0,\\\end{alignedat}}\right.

Thus

(-1)x=-x

and it can be shown that for all integers

m,n\in \mathbb {Z}

and all

x\in X,

(m+n)x=mx+nx

and

-(nx)=(-n)x=n(-x).

This definition of scalar multiplication makes the cyclic subgroup

\mathbb {Z} x

of

X

into a left

\mathbb {Z}

-module; if

X

is commutative, then it also makes

X

into a left

\mathbb {Z}

-module.

Homogeneity over the integers

If $f:X\to Y$ is an additive map between additive groups then $f(0)=0$ and for all $x\in X,$ $f(-x)=-f(x)$ (where negation denotes the additive inverse) and^{[proof 1]}

f(nx)=nf(x)\quad {\text{ for all }}n\in \mathbb {Z} .

Consequently,

f(x-y)=f(x)-f(y)

for all

x,y\in X

(where by definition,

x-y:=x+(-y)

).

In other words, every additive map is homogeneous over the integers. Consequently, every additive map between abelian groups is a homomorphism of $\mathbb {Z}$ -modules.

Homomorphism of $\mathbb {Q}$ -modules

If the additive abelian groups $X$ and $Y$ are also a unital modules over the rationals $\mathbb {Q}$ (such as real or complex vector spaces) then an additive map $f:X\to Y$ satisfies:^{[proof 2]}

f(qx)=qf(x)\quad {\text{ for all }}q\in \mathbb {Q} {\text{ and }}x\in X.

In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital

\mathbb {Q}

-modules is a homomorphism of

\mathbb {Q}

-modules.

Despite being homogeneous over $\mathbb {Q} ,$ as described in the article on Cauchy's functional equation, even when $X=Y=\mathbb {R} ,$ it is nevertheless still possible for the additive function $f:\mathbb {R} \to \mathbb {R}$ to not be homogeneous over the real numbers; said differently, there exist additive maps $f:\mathbb {R} \to \mathbb {R}$ that are not of the form $f(x)=s_{0}x$ for some constant $s_{0}\in \mathbb {R} .$ In particular, there exist additive maps that are not linear maps.

Notes edit

^ Leslie Hogben (2013), Handbook of Linear Algebra (3 ed.), CRC Press, pp. 30–8, ISBN 9781498785600
^ N. Bourbaki (1989), Algebra Chapters 1–3, Springer, p. 243

Proofs

^ $f(0)=f(0+0)=f(0)+f(0)$ so adding $-f(0)$ to both sides proves that $f(0)=0.$ If $x\in X$ 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 $n\in \mathbb {N}$ 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 $n<0$ ). $\blacksquare$
^ Let $x\in X$ and $q={\frac {m}{n}}\in \mathbb {Q}$ where $m,n\in \mathbb {Z}$ and $n>0.$ Let $y:={\frac {1}{n}}x.$ Then $ny=n\left({\frac {1}{n}}x\right)=\left(n{\frac {1}{n}}\right)x=(1)x=x,$ which implies $f(x)=f(ny)=nf(y)=nf\left({\frac {1}{n}}x\right)$ so that multiplying both sides by ${\frac {1}{n}}$ proves that $f\left({\frac {1}{n}}x\right)={\frac {1}{n}}f(x).$ Consequently, $f(qx)=f\left({\frac {m}{n}}x\right)=mf\left({\frac {1}{n}}x\right)=m\left({\frac {1}{n}}f(x)\right)=qf(x).$ $\blacksquare$

References edit

Roger C. Lyndon; Paul E. Schupp (2001), Combinatorial Group Theory, Springer

[1] Leslie Hogben (2013), Handbook of Linear Algebra (3 ed.), CRC Press, pp. 30–8, ISBN 9781498785600

[2] N. Bourbaki (1989), Algebra Chapters 1–3, Springer, p. 243

[3] $f(0)=f(0+0)=f(0)+f(0)$ so adding $-f(0)$ to both sides proves that $f(0)=0.$ If $x\in X$ 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 $n\in \mathbb {N}$ 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 $n<0$ ). $\blacksquare$

[4] Let $x\in X$ and $q={\frac {m}{n}}\in \mathbb {Q}$ where $m,n\in \mathbb {Z}$ and $n>0.$ Let $y:={\frac {1}{n}}x.$ Then $ny=n\left({\frac {1}{n}}x\right)=\left(n{\frac {1}{n}}\right)x=(1)x=x,$ which implies $f(x)=f(ny)=nf(y)=nf\left({\frac {1}{n}}x\right)$ so that multiplying both sides by ${\frac {1}{n}}$ proves that $f\left({\frac {1}{n}}x\right)={\frac {1}{n}}f(x).$ Consequently, $f(qx)=f\left({\frac {m}{n}}x\right)=mf\left({\frac {1}{n}}x\right)=m\left({\frac {1}{n}}f(x)\right)=qf(x).$ $\blacksquare$

[1]

[2]

[proof 1]

[proof 2]

Summary

Examples edit

Properties edit

See also edit

Notes edit

References edit