Free vector space of a finite set is isomorphic to its function space












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Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?






linear-algebra functional-analysis







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  • $begingroup$
    $X$ is finite, so it is not infinite dimensional.
    $endgroup$
    – Randall
    Jan 16 at 3:22






  • 1




    $begingroup$
    An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
    $endgroup$
    – Randall
    Jan 16 at 3:24


















0












$begingroup$


Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?






linear-algebra functional-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    $X$ is finite, so it is not infinite dimensional.
    $endgroup$
    – Randall
    Jan 16 at 3:22






  • 1




    $begingroup$
    An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
    $endgroup$
    – Randall
    Jan 16 at 3:24
















0












0








0





$begingroup$


Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?






linear-algebra functional-analysis







share|cite|improve this question









$endgroup$




Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?






linear-algebra functional-analysis




linear-algebra functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 3:21









David FengDavid Feng

979




979












  • $begingroup$
    $X$ is finite, so it is not infinite dimensional.
    $endgroup$
    – Randall
    Jan 16 at 3:22






  • 1




    $begingroup$
    An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
    $endgroup$
    – Randall
    Jan 16 at 3:24




















  • $begingroup$
    $X$ is finite, so it is not infinite dimensional.
    $endgroup$
    – Randall
    Jan 16 at 3:22






  • 1




    $begingroup$
    An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
    $endgroup$
    – Randall
    Jan 16 at 3:24


















$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22




$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22




1




1




$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24






$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24












1 Answer
1






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oldest

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$begingroup$

Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.



So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso






share|cite|improve this answer









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    $begingroup$

    Consider the map
    $$varphi : F(X,mathbb{C}) to mathbb{C}X\
    f mapsto sum_{xin X} f(x)x$$
    we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
    let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
    Now for injectivity:
    since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.



    So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
    And so $varphi$ is an iso






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Consider the map
      $$varphi : F(X,mathbb{C}) to mathbb{C}X\
      f mapsto sum_{xin X} f(x)x$$
      we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
      let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
      Now for injectivity:
      since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.



      So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
      And so $varphi$ is an iso






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Consider the map
        $$varphi : F(X,mathbb{C}) to mathbb{C}X\
        f mapsto sum_{xin X} f(x)x$$
        we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
        let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
        Now for injectivity:
        since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.



        So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
        And so $varphi$ is an iso






        share|cite|improve this answer









        $endgroup$



        Consider the map
        $$varphi : F(X,mathbb{C}) to mathbb{C}X\
        f mapsto sum_{xin X} f(x)x$$
        we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
        let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
        Now for injectivity:
        since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.



        So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
        And so $varphi$ is an iso







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 8:13









        EnkiduEnkidu

        1,36719




        1,36719






























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