In algebra , an additive map ,
Z
{\displaystyle Z}
additive function is a function
f
{\displaystyle f}
[ 1]
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
{\displaystyle f(x+y)=f(x)+f(y)}
x
{\displaystyle x}
y
{\displaystyle y}
domain of
f
.
{\displaystyle f.}
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
Z
{\displaystyle \mathbb {Z} }
module homomorphism . Since an abelian group is a
Z
{\displaystyle \mathbb {Z} }
module , it may be defined as a group homomorphism between abelian groups.
A map
V
×
W
→
X
{\displaystyle V\times W\to X}
bi-additive map or a
Z
{\displaystyle \mathbb {Z} }
[ 2]
Examples
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
{\displaystyle f}
g
{\displaystyle g}
f
+
g
{\displaystyle f+g}
pointwise ) is additive.
Properties
Suppose that
X
{\displaystyle X}
0
{\displaystyle 0}
x
∈
X
{\displaystyle x\in X}
−
x
.
{\displaystyle -x.}
x
∈
X
,
{\displaystyle x\in X,}
n
x
:=
0
when
n
=
0
,
(
−
1
)
x
:=
−
x
when
n
=
−
1
,
n
x
:=
x
+
⋯
+
x
(
n
-terms)
when
n
>
0
,
n
x
:=
(
−
n
)
(
−
x
)
when
n
<
0.
{\displaystyle {\begin{alignedat}{9}nx&:=0&&{\text{ when }}n=0,\\(-1)x&:=-x&&{\text{ when }}n=-1,\\nx&:=x+\cdots +x~~{\text{(}}n{\text{-terms) }}&&{\text{ when }}n>0,\\nx&:=(-n)(-x)&&{\text{ when }}n<0.\end{alignedat}}}
m
,
n
∈
Z
{\displaystyle m,n\in \mathbb {Z} }
x
∈
X
,
{\displaystyle x\in X,}
(
m
+
n
)
x
=
m
x
+
n
x
{\displaystyle (m+n)x=mx+nx}
n
x
{\displaystyle nx}
(
−
n
)
x
;
{\displaystyle (-n)x;}
−
(
n
x
)
=
(
−
n
)
x
.
{\displaystyle -(nx)=(-n)x.}
If
f
:
X
→
Y
{\displaystyle f:X\to Y}
f
(
0
)
=
0
{\displaystyle f(0)=0}
x
∈
X
,
{\displaystyle x\in X,}
f
(
−
x
)
=
−
f
(
x
)
{\displaystyle f(-x)=-f(x)}
[ proof 1]
f
(
n
x
)
=
n
f
(
x
)
for all
n
∈
Z
.
{\displaystyle f(nx)=nf(x)\quad {\text{ for all }}n\in \mathbb {Z} .}
If the additive groups
X
{\displaystyle X}
Y
{\displaystyle Y}
modules over the rationals
Q
{\displaystyle \mathbb {Q} }
vector space ) then[ proof 2]
f
(
q
x
)
=
q
f
(
x
)
for all
q
∈
Q
and
x
∈
X
.
{\displaystyle f(qx)=qf(x)\quad {\text{ for all }}q\in \mathbb {Q} {\text{ and }}x\in X.}
See also
Notes
^ 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
)
{\displaystyle f(0)=f(0+0)=f(0)+f(0)}
−
f
(
0
)
{\displaystyle -f(0)}
f
(
0
)
=
0.
{\displaystyle f(0)=0.}
x
∈
X
{\displaystyle x\in X}
0
=
f
(
0
)
=
f
(
x
+
(
−
x
)
)
=
f
(
x
)
+
f
(
−
x
)
{\displaystyle 0=f(0)=f(x+(-x))=f(x)+f(-x)}
f
(
−
x
)
=
−
f
(
x
)
{\displaystyle f(-x)=-f(x)}
(
−
1
)
f
(
x
)
:=
−
f
(
x
)
.
{\displaystyle (-1)f(x):=-f(x).}
n
∈
N
{\displaystyle n\in \mathbb {N} }
f
(
n
x
)
=
n
f
(
x
)
{\displaystyle f(nx)=nf(x)}
n
f
(
x
)
{\displaystyle nf(x)}
n
(
−
f
(
x
)
)
,
{\displaystyle n(-f(x)),}
f
(
(
−
n
)
x
)
=
f
(
n
(
−
x
)
)
=
n
f
(
−
x
)
=
n
(
−
f
(
x
)
)
=
−
(
n
f
(
x
)
)
=
(
−
n
)
f
(
x
)
{\displaystyle f((-n)x)=f(n(-x))=nf(-x)=n(-f(x))=-(nf(x))=(-n)f(x)}
f
(
n
x
)
=
n
f
(
x
)
{\displaystyle f(nx)=nf(x)}
n
<
0
{\displaystyle n<0}
◼
{\displaystyle \blacksquare }
^ Let
x
∈
X
{\displaystyle x\in X}
q
=
m
n
∈
Q
{\displaystyle q={\frac {m}{n}}\in \mathbb {Q} }
m
,
n
∈
Z
{\displaystyle m,n\in \mathbb {Z} }
n
>
0.
{\displaystyle n>0.}
y
:=
1
n
x
.
{\displaystyle y:={\frac {1}{n}}x.}
n
y
=
n
(
1
n
x
)
=
(
n
1
n
)
x
=
(
1
)
x
=
x
,
{\displaystyle ny=n\left({\frac {1}{n}}x\right)=\left(n{\frac {1}{n}}\right)x=(1)x=x,}
f
(
x
)
=
f
(
n
y
)
=
n
f
(
y
)
=
n
f
(
1
n
x
)
{\displaystyle f(x)=f(ny)=nf(y)=nf\left({\frac {1}{n}}x\right)}
1
n
{\displaystyle {\frac {1}{n}}}
f
(
1
n
x
)
=
1
n
f
(
x
)
.
{\displaystyle f\left({\frac {1}{n}}x\right)={\frac {1}{n}}f(x).}
f
(
q
x
)
=
f
(
m
n
x
)
=
m
f
(
1
n
x
)
=
m
(
1
n
f
(
x
)
)
=
q
f
(
x
)
.
{\displaystyle 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).}
◼
{\displaystyle \blacksquare }
References
