Function with finite domain can be linear?












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$begingroup$


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







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$endgroup$












  • $begingroup$
    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


















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 14:36









Yanko

7,8801830




7,8801830










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




















  • $begingroup$
    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












2 Answers
2






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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.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    No because in general $f(ax)$ is not defined and cannot equal $af(x)$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      3












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 14:40









          Clive NewsteadClive Newstead

          51.9k474136




          51.9k474136























              3












              $begingroup$

              No because in general $f(ax)$ is not defined and cannot equal $af(x)$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                No because in general $f(ax)$ is not defined and cannot equal $af(x)$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  No because in general $f(ax)$ is not defined and cannot equal $af(x)$.






                  share|cite|improve this answer









                  $endgroup$



                  No because in general $f(ax)$ is not defined and cannot equal $af(x)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 25 at 14:40









                  Yves DaoustYves Daoust

                  131k676229




                  131k676229






























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