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Finding a basis of an infinite dimensional vector space with a given vector

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Finding a basis of an infinite dimensional vector space with a given vector


An orthonormal set cannot be a basis in an infinite dimension vector space?Finding a basis of an infinite-dimensional vector space?Extension of linearly independent set to a basis in an infinite dimensional vector spacerepresentation of a vector in an $n$-dimensional vector space $X$ is uniqueAn explicit basis for an infinite dimensional vector space.Number of two dimensional sub spaces of a vector space over a finite field.Require a basis such that a given vector has components (1,0,0,…,0)Why does dual basis not span the dual space when the given vector space V is infinite dimensional?Proving a vector space is infinite-dimensionalBasis of infinite Vector Space $mathbb R^{infty}$













2












$begingroup$


If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






linear-algebra set-theory axiom-of-choice







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



    First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



    My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



    P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






    linear-algebra set-theory axiom-of-choice







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



      First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



      My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



      P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






      linear-algebra set-theory axiom-of-choice







      share|cite|improve this question











      $endgroup$




      If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:



      First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.



      My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?



      P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.






      linear-algebra set-theory axiom-of-choice




      linear-algebra set-theory axiom-of-choice






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Robert Shore

      1,32815




      1,32815










      asked 2 hours ago









      JustDroppedInJustDroppedIn

      2,054420




      2,054420






















          2 Answers
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          active

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          3












          $begingroup$

          Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
            $$
            v = a + k_2a_2 + cdots + k_na_n .
            $$



            Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

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              3












              $begingroup$

              Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.






                  share|cite|improve this answer











                  $endgroup$



                  Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  Robert ShoreRobert Shore

                  1,32815




                  1,32815























                      1












                      $begingroup$

                      Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                      $$
                      v = a + k_2a_2 + cdots + k_na_n .
                      $$



                      Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                        $$
                        v = a + k_2a_2 + cdots + k_na_n .
                        $$



                        Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                          $$
                          v = a + k_2a_2 + cdots + k_na_n .
                          $$



                          Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.






                          share|cite|improve this answer









                          $endgroup$



                          Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
                          $$
                          v = a + k_2a_2 + cdots + k_na_n .
                          $$



                          Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Ethan BolkerEthan Bolker

                          43.5k551116




                          43.5k551116






























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