Mixing
Function with finite domain can be linear?
$begingroup$
Consider a function $f:mathcal{X}rightarrow mathcal{Y}$ prescribed by $f(x)=3x$,
where $mathcal{X}={1,2,3}$ and $mathcal{Y}equiv {3,6,9}$.
Can we say that $f$ is linear? My confusion stems from the finite domain.
linear-algebra functions
edited Jan 25 at 14:36
Yanko
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
$endgroup$
add a comment |
$begingroup$
Consider a function $f:mathcal{X}rightarrow mathcal{Y}$ prescribed by $f(x)=3x$,
where $mathcal{X}={1,2,3}$ and $mathcal{Y}equiv {3,6,9}$.
Can we say that $f$ is linear? My confusion stems from the finite domain.
linear-algebra functions
edited Jan 25 at 14:36
Yanko
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
$endgroup$
$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35
add a comment |
$begingroup$
Consider a function $f:mathcal{X}rightarrow mathcal{Y}$ prescribed by $f(x)=3x$,
where $mathcal{X}={1,2,3}$ and $mathcal{Y}equiv {3,6,9}$.
Can we say that $f$ is linear? My confusion stems from the finite domain.
linear-algebra functions
edited Jan 25 at 14:36
Yanko
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
$endgroup$
Consider a function $f:mathcal{X}rightarrow mathcal{Y}$ prescribed by $f(x)=3x$,
where $mathcal{X}={1,2,3}$ and $mathcal{Y}equiv {3,6,9}$.
Can we say that $f$ is linear? My confusion stems from the finite domain.
linear-algebra functions
linear-algebra functions
edited Jan 25 at 14:36
Yanko
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
edited Jan 25 at 14:36
Yanko
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
edited Jan 25 at 14:36
Yanko
7,8801830
edited Jan 25 at 14:36
Yanko
7,8801830
edited Jan 25 at 14:36
Yanko
7,8801830
7,8801830
asked Jan 25 at 14:34
STFSTF
541422
asked Jan 25 at 14:34
STFSTF
541422
asked Jan 25 at 14:34
STFSTF
541422
541422
$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35
add a comment |
$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35
$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35
$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35
add a comment |
2 Answers
2
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votes
$begingroup$
Typically we only ask if a function is linear if it is a function between vector spaces (a.k.a. linear spaces), or some other suitably structured set for which a notion of 'linearity' makes sense, such as a module over a ring.
A function $f$ between two such structures is then linear if it preserves the linear structure: this usually amounts to saying something like $f(ax+by) = af(x) +bf(y)$ for all vectors $x,y$ and all scalars $a,b$.
In this case, $mathcal{X}$ and $mathcal{Y}$ are just sets with three elements. They are certainly not vector spaces or modules over a ring, so it doesn't really make sense to ask if $f$ is linear.
As Yanko says in a comment, though, we can embed $mathcal{X}$ and $mathcal{Y}$ in the vector space $mathbb{R}$, and then $f$ is the (co)restriction of a linear function $mathbb{R} to mathbb{R}$ to a function $mathcal{X} to mathcal{Y}$. However, there are also functions $mathbb{R} to mathbb{R}$ which are not linear, but whose (co)restriction to $mathcal{X}$ and $mathcal{Y}$ is the function $x mapsto 3x$. One example is $x mapsto x^3-6x^2+14x-6$.
So is $f$ linear? Not really.
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
add a comment |
$begingroup$
No because in general $f(ax)$ is not defined and cannot equal $af(x)$.
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Typically we only ask if a function is linear if it is a function between vector spaces (a.k.a. linear spaces), or some other suitably structured set for which a notion of 'linearity' makes sense, such as a module over a ring.
A function $f$ between two such structures is then linear if it preserves the linear structure: this usually amounts to saying something like $f(ax+by) = af(x) +bf(y)$ for all vectors $x,y$ and all scalars $a,b$.
In this case, $mathcal{X}$ and $mathcal{Y}$ are just sets with three elements. They are certainly not vector spaces or modules over a ring, so it doesn't really make sense to ask if $f$ is linear.
As Yanko says in a comment, though, we can embed $mathcal{X}$ and $mathcal{Y}$ in the vector space $mathbb{R}$, and then $f$ is the (co)restriction of a linear function $mathbb{R} to mathbb{R}$ to a function $mathcal{X} to mathcal{Y}$. However, there are also functions $mathbb{R} to mathbb{R}$ which are not linear, but whose (co)restriction to $mathcal{X}$ and $mathcal{Y}$ is the function $x mapsto 3x$. One example is $x mapsto x^3-6x^2+14x-6$.
So is $f$ linear? Not really.
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
add a comment |
$begingroup$
Typically we only ask if a function is linear if it is a function between vector spaces (a.k.a. linear spaces), or some other suitably structured set for which a notion of 'linearity' makes sense, such as a module over a ring.
A function $f$ between two such structures is then linear if it preserves the linear structure: this usually amounts to saying something like $f(ax+by) = af(x) +bf(y)$ for all vectors $x,y$ and all scalars $a,b$.
In this case, $mathcal{X}$ and $mathcal{Y}$ are just sets with three elements. They are certainly not vector spaces or modules over a ring, so it doesn't really make sense to ask if $f$ is linear.
As Yanko says in a comment, though, we can embed $mathcal{X}$ and $mathcal{Y}$ in the vector space $mathbb{R}$, and then $f$ is the (co)restriction of a linear function $mathbb{R} to mathbb{R}$ to a function $mathcal{X} to mathcal{Y}$. However, there are also functions $mathbb{R} to mathbb{R}$ which are not linear, but whose (co)restriction to $mathcal{X}$ and $mathcal{Y}$ is the function $x mapsto 3x$. One example is $x mapsto x^3-6x^2+14x-6$.
So is $f$ linear? Not really.
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
add a comment |
$begingroup$
Typically we only ask if a function is linear if it is a function between vector spaces (a.k.a. linear spaces), or some other suitably structured set for which a notion of 'linearity' makes sense, such as a module over a ring.
A function $f$ between two such structures is then linear if it preserves the linear structure: this usually amounts to saying something like $f(ax+by) = af(x) +bf(y)$ for all vectors $x,y$ and all scalars $a,b$.
In this case, $mathcal{X}$ and $mathcal{Y}$ are just sets with three elements. They are certainly not vector spaces or modules over a ring, so it doesn't really make sense to ask if $f$ is linear.
As Yanko says in a comment, though, we can embed $mathcal{X}$ and $mathcal{Y}$ in the vector space $mathbb{R}$, and then $f$ is the (co)restriction of a linear function $mathbb{R} to mathbb{R}$ to a function $mathcal{X} to mathcal{Y}$. However, there are also functions $mathbb{R} to mathbb{R}$ which are not linear, but whose (co)restriction to $mathcal{X}$ and $mathcal{Y}$ is the function $x mapsto 3x$. One example is $x mapsto x^3-6x^2+14x-6$.
So is $f$ linear? Not really.
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
$endgroup$
Typically we only ask if a function is linear if it is a function between vector spaces (a.k.a. linear spaces), or some other suitably structured set for which a notion of 'linearity' makes sense, such as a module over a ring.
A function $f$ between two such structures is then linear if it preserves the linear structure: this usually amounts to saying something like $f(ax+by) = af(x) +bf(y)$ for all vectors $x,y$ and all scalars $a,b$.
In this case, $mathcal{X}$ and $mathcal{Y}$ are just sets with three elements. They are certainly not vector spaces or modules over a ring, so it doesn't really make sense to ask if $f$ is linear.
As Yanko says in a comment, though, we can embed $mathcal{X}$ and $mathcal{Y}$ in the vector space $mathbb{R}$, and then $f$ is the (co)restriction of a linear function $mathbb{R} to mathbb{R}$ to a function $mathcal{X} to mathcal{Y}$. However, there are also functions $mathbb{R} to mathbb{R}$ which are not linear, but whose (co)restriction to $mathcal{X}$ and $mathcal{Y}$ is the function $x mapsto 3x$. One example is $x mapsto x^3-6x^2+14x-6$.
So is $f$ linear? Not really.
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
answered Jan 25 at 14:40
Clive NewsteadClive Newstead
51.9k474136
51.9k474136
add a comment |
add a comment |
$begingroup$
No because in general $f(ax)$ is not defined and cannot equal $af(x)$.
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
No because in general $f(ax)$ is not defined and cannot equal $af(x)$.
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
No because in general $f(ax)$ is not defined and cannot equal $af(x)$.
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
$endgroup$
No because in general $f(ax)$ is not defined and cannot equal $af(x)$.
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
answered Jan 25 at 14:40
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
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$begingroup$
You can say that $f$ is the restriction of the linear function $f(x)=3x$ to the set ${1,2,3}$. This is uncommon though.
$endgroup$
– Yanko
Jan 25 at 14:35