Basis for vector space $F(X,V)$ of functions mapping any arbitrary set to a vector space












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How would I go about determining the basis for the vector space $F(X,V)$ of all functions mapping an element of any arbitrary set $X$ to some vector space $V$ with the usual definition of addition and scalar multiplication on functions, should I already know the basis for that vector space?



I tried to extend the notion of the basis of the vector space of linear maps $F(X,F)$ from some set $X={a_0,...,a_n}$ to an arbitrary field where the basis was made up of functions defined as $f_i(a_j)=1$ when $i=j$ and $0$ otherwise. However I have had no such luck developing the idea for $F(X,V)$.






linear-algebra vector-spaces hamel-basis







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    0












    $begingroup$


    How would I go about determining the basis for the vector space $F(X,V)$ of all functions mapping an element of any arbitrary set $X$ to some vector space $V$ with the usual definition of addition and scalar multiplication on functions, should I already know the basis for that vector space?



    I tried to extend the notion of the basis of the vector space of linear maps $F(X,F)$ from some set $X={a_0,...,a_n}$ to an arbitrary field where the basis was made up of functions defined as $f_i(a_j)=1$ when $i=j$ and $0$ otherwise. However I have had no such luck developing the idea for $F(X,V)$.






    linear-algebra vector-spaces hamel-basis







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      How would I go about determining the basis for the vector space $F(X,V)$ of all functions mapping an element of any arbitrary set $X$ to some vector space $V$ with the usual definition of addition and scalar multiplication on functions, should I already know the basis for that vector space?



      I tried to extend the notion of the basis of the vector space of linear maps $F(X,F)$ from some set $X={a_0,...,a_n}$ to an arbitrary field where the basis was made up of functions defined as $f_i(a_j)=1$ when $i=j$ and $0$ otherwise. However I have had no such luck developing the idea for $F(X,V)$.






      linear-algebra vector-spaces hamel-basis







      share|cite|improve this question











      $endgroup$




      How would I go about determining the basis for the vector space $F(X,V)$ of all functions mapping an element of any arbitrary set $X$ to some vector space $V$ with the usual definition of addition and scalar multiplication on functions, should I already know the basis for that vector space?



      I tried to extend the notion of the basis of the vector space of linear maps $F(X,F)$ from some set $X={a_0,...,a_n}$ to an arbitrary field where the basis was made up of functions defined as $f_i(a_j)=1$ when $i=j$ and $0$ otherwise. However I have had no such luck developing the idea for $F(X,V)$.






      linear-algebra vector-spaces hamel-basis




      linear-algebra vector-spaces hamel-basis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 10:48









      José Carlos Santos

      158k22126228




      158k22126228










      asked Dec 8 '17 at 15:00









      Michael ConnorMichael Connor

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      404212






















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          $begingroup$

          Here's an answer, assuming that $V$ is finite-dimensional. Let $B$ be a basis of $V$. For each $xin X$ and each $win B$, let $f_{x,w}in F(X,V)$ be the function defined by$$f_{x,w}(y)=begin{cases}w&text{ if }y=x\0&text{ otherwise.}end{cases}$$Then ${f_{x,w},|,xin Xwedge win B}$ is a basis of $F(X,V)$.






          share|cite|improve this answer









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          • $begingroup$
            This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
            $endgroup$
            – Arnaud D.
            Dec 8 '17 at 15:30











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Here's an answer, assuming that $V$ is finite-dimensional. Let $B$ be a basis of $V$. For each $xin X$ and each $win B$, let $f_{x,w}in F(X,V)$ be the function defined by$$f_{x,w}(y)=begin{cases}w&text{ if }y=x\0&text{ otherwise.}end{cases}$$Then ${f_{x,w},|,xin Xwedge win B}$ is a basis of $F(X,V)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
            $endgroup$
            – Arnaud D.
            Dec 8 '17 at 15:30
















          1












          $begingroup$

          Here's an answer, assuming that $V$ is finite-dimensional. Let $B$ be a basis of $V$. For each $xin X$ and each $win B$, let $f_{x,w}in F(X,V)$ be the function defined by$$f_{x,w}(y)=begin{cases}w&text{ if }y=x\0&text{ otherwise.}end{cases}$$Then ${f_{x,w},|,xin Xwedge win B}$ is a basis of $F(X,V)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
            $endgroup$
            – Arnaud D.
            Dec 8 '17 at 15:30














          1












          1








          1





          $begingroup$

          Here's an answer, assuming that $V$ is finite-dimensional. Let $B$ be a basis of $V$. For each $xin X$ and each $win B$, let $f_{x,w}in F(X,V)$ be the function defined by$$f_{x,w}(y)=begin{cases}w&text{ if }y=x\0&text{ otherwise.}end{cases}$$Then ${f_{x,w},|,xin Xwedge win B}$ is a basis of $F(X,V)$.






          share|cite|improve this answer









          $endgroup$



          Here's an answer, assuming that $V$ is finite-dimensional. Let $B$ be a basis of $V$. For each $xin X$ and each $win B$, let $f_{x,w}in F(X,V)$ be the function defined by$$f_{x,w}(y)=begin{cases}w&text{ if }y=x\0&text{ otherwise.}end{cases}$$Then ${f_{x,w},|,xin Xwedge win B}$ is a basis of $F(X,V)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '17 at 15:06









          José Carlos SantosJosé Carlos Santos

          158k22126228




          158k22126228












          • $begingroup$
            This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
            $endgroup$
            – Arnaud D.
            Dec 8 '17 at 15:30


















          • $begingroup$
            This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
            $endgroup$
            – Arnaud D.
            Dec 8 '17 at 15:30
















          $begingroup$
          This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
          $endgroup$
          – Arnaud D.
          Dec 8 '17 at 15:30




          $begingroup$
          This does'nt work if $X$ is infinite : for example a non-zero constant function can't be obtained as a finite linear combination of $f_{x,w}$, as it would be zero for some $y$. This is also true for the case $V=F$ mentioned in the OP.
          $endgroup$
          – Arnaud D.
          Dec 8 '17 at 15:30


















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