expanding a independent set to a basis












0












$begingroup$


I doubt that I misunderstand this part:



please clarify me:



if $V$ is a finite-dimensional vector space and we have an independent set like $S$ then we can expand $S$ to a basis for the vector space easily.



if $V$ is an infinite-dimensional vector space and we have an independent set $S$ then again we can expand $S$ to a basis for the vector space but this time the procedure is more complicated and for proving this we must use Zorn's lemma



are this two statements correct?






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    yes the are correct
    $endgroup$
    – Enkidu
    Dec 5 '18 at 13:10






  • 2




    $begingroup$
    It's quite correct. As a consequence, every vector space has a basis.
    $endgroup$
    – Bernard
    Dec 5 '18 at 13:10










  • $begingroup$
    How would you use Zorn's lemma?
    $endgroup$
    – nafhgood
    Dec 5 '18 at 13:13
















0












$begingroup$


I doubt that I misunderstand this part:



please clarify me:



if $V$ is a finite-dimensional vector space and we have an independent set like $S$ then we can expand $S$ to a basis for the vector space easily.



if $V$ is an infinite-dimensional vector space and we have an independent set $S$ then again we can expand $S$ to a basis for the vector space but this time the procedure is more complicated and for proving this we must use Zorn's lemma



are this two statements correct?






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    yes the are correct
    $endgroup$
    – Enkidu
    Dec 5 '18 at 13:10






  • 2




    $begingroup$
    It's quite correct. As a consequence, every vector space has a basis.
    $endgroup$
    – Bernard
    Dec 5 '18 at 13:10










  • $begingroup$
    How would you use Zorn's lemma?
    $endgroup$
    – nafhgood
    Dec 5 '18 at 13:13














0












0








0





$begingroup$


I doubt that I misunderstand this part:



please clarify me:



if $V$ is a finite-dimensional vector space and we have an independent set like $S$ then we can expand $S$ to a basis for the vector space easily.



if $V$ is an infinite-dimensional vector space and we have an independent set $S$ then again we can expand $S$ to a basis for the vector space but this time the procedure is more complicated and for proving this we must use Zorn's lemma



are this two statements correct?






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$




I doubt that I misunderstand this part:



please clarify me:



if $V$ is a finite-dimensional vector space and we have an independent set like $S$ then we can expand $S$ to a basis for the vector space easily.



if $V$ is an infinite-dimensional vector space and we have an independent set $S$ then again we can expand $S$ to a basis for the vector space but this time the procedure is more complicated and for proving this we must use Zorn's lemma



are this two statements correct?






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 13:00









jack.mathjack.math

616




616












  • $begingroup$
    yes the are correct
    $endgroup$
    – Enkidu
    Dec 5 '18 at 13:10






  • 2




    $begingroup$
    It's quite correct. As a consequence, every vector space has a basis.
    $endgroup$
    – Bernard
    Dec 5 '18 at 13:10










  • $begingroup$
    How would you use Zorn's lemma?
    $endgroup$
    – nafhgood
    Dec 5 '18 at 13:13


















  • $begingroup$
    yes the are correct
    $endgroup$
    – Enkidu
    Dec 5 '18 at 13:10






  • 2




    $begingroup$
    It's quite correct. As a consequence, every vector space has a basis.
    $endgroup$
    – Bernard
    Dec 5 '18 at 13:10










  • $begingroup$
    How would you use Zorn's lemma?
    $endgroup$
    – nafhgood
    Dec 5 '18 at 13:13
















$begingroup$
yes the are correct
$endgroup$
– Enkidu
Dec 5 '18 at 13:10




$begingroup$
yes the are correct
$endgroup$
– Enkidu
Dec 5 '18 at 13:10




2




2




$begingroup$
It's quite correct. As a consequence, every vector space has a basis.
$endgroup$
– Bernard
Dec 5 '18 at 13:10




$begingroup$
It's quite correct. As a consequence, every vector space has a basis.
$endgroup$
– Bernard
Dec 5 '18 at 13:10












$begingroup$
How would you use Zorn's lemma?
$endgroup$
– nafhgood
Dec 5 '18 at 13:13




$begingroup$
How would you use Zorn's lemma?
$endgroup$
– nafhgood
Dec 5 '18 at 13:13










2 Answers
2






active

oldest

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1












$begingroup$

@mathnoob Let $mathcal{P}={Lsupset S|Ltext{ is linearly independent}}$.This is nonempty since $Sin mathcal{P}$. Then $(mathcal{P},subset)$ is a poset with set inclusion. Use zorn's lemma on this to get a maximal linearly independent set $B$ containing $S$. Then $B$ is an extension of $S$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    In case the vector space $V$ is finite dimensional it has a finite spanning set say $X={x_1,dots,x_n}$. The independent set $S$ is going to have cardinality atmost $n$(see Steinitz exchange theorem). Now keep adding vectors from $X$ to $S$ which are not there in the span of the vectors already there. For instance if $x_1$ is there in span($S$) ignore otherwise adjoin to get $Scup{x_1}$. Now if $x_2$ is there in span($Scup{x_1}$) or span($S$)(by whatever happened in previous step) then ignore otherwise adjoin to get $Scup{x_1,x_2}$ or $Scup{x_2}$. Continue like this for $n$ steps to get a linearly independent set (since none of the vectors are in the span of the previous ones) containing $S$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      @mathnoob Let $mathcal{P}={Lsupset S|Ltext{ is linearly independent}}$.This is nonempty since $Sin mathcal{P}$. Then $(mathcal{P},subset)$ is a poset with set inclusion. Use zorn's lemma on this to get a maximal linearly independent set $B$ containing $S$. Then $B$ is an extension of $S$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        @mathnoob Let $mathcal{P}={Lsupset S|Ltext{ is linearly independent}}$.This is nonempty since $Sin mathcal{P}$. Then $(mathcal{P},subset)$ is a poset with set inclusion. Use zorn's lemma on this to get a maximal linearly independent set $B$ containing $S$. Then $B$ is an extension of $S$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          @mathnoob Let $mathcal{P}={Lsupset S|Ltext{ is linearly independent}}$.This is nonempty since $Sin mathcal{P}$. Then $(mathcal{P},subset)$ is a poset with set inclusion. Use zorn's lemma on this to get a maximal linearly independent set $B$ containing $S$. Then $B$ is an extension of $S$.






          share|cite|improve this answer









          $endgroup$



          @mathnoob Let $mathcal{P}={Lsupset S|Ltext{ is linearly independent}}$.This is nonempty since $Sin mathcal{P}$. Then $(mathcal{P},subset)$ is a poset with set inclusion. Use zorn's lemma on this to get a maximal linearly independent set $B$ containing $S$. Then $B$ is an extension of $S$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 13:26









          Arpan DasArpan Das

          837




          837























              1












              $begingroup$

              In case the vector space $V$ is finite dimensional it has a finite spanning set say $X={x_1,dots,x_n}$. The independent set $S$ is going to have cardinality atmost $n$(see Steinitz exchange theorem). Now keep adding vectors from $X$ to $S$ which are not there in the span of the vectors already there. For instance if $x_1$ is there in span($S$) ignore otherwise adjoin to get $Scup{x_1}$. Now if $x_2$ is there in span($Scup{x_1}$) or span($S$)(by whatever happened in previous step) then ignore otherwise adjoin to get $Scup{x_1,x_2}$ or $Scup{x_2}$. Continue like this for $n$ steps to get a linearly independent set (since none of the vectors are in the span of the previous ones) containing $S$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In case the vector space $V$ is finite dimensional it has a finite spanning set say $X={x_1,dots,x_n}$. The independent set $S$ is going to have cardinality atmost $n$(see Steinitz exchange theorem). Now keep adding vectors from $X$ to $S$ which are not there in the span of the vectors already there. For instance if $x_1$ is there in span($S$) ignore otherwise adjoin to get $Scup{x_1}$. Now if $x_2$ is there in span($Scup{x_1}$) or span($S$)(by whatever happened in previous step) then ignore otherwise adjoin to get $Scup{x_1,x_2}$ or $Scup{x_2}$. Continue like this for $n$ steps to get a linearly independent set (since none of the vectors are in the span of the previous ones) containing $S$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In case the vector space $V$ is finite dimensional it has a finite spanning set say $X={x_1,dots,x_n}$. The independent set $S$ is going to have cardinality atmost $n$(see Steinitz exchange theorem). Now keep adding vectors from $X$ to $S$ which are not there in the span of the vectors already there. For instance if $x_1$ is there in span($S$) ignore otherwise adjoin to get $Scup{x_1}$. Now if $x_2$ is there in span($Scup{x_1}$) or span($S$)(by whatever happened in previous step) then ignore otherwise adjoin to get $Scup{x_1,x_2}$ or $Scup{x_2}$. Continue like this for $n$ steps to get a linearly independent set (since none of the vectors are in the span of the previous ones) containing $S$.






                  share|cite|improve this answer









                  $endgroup$



                  In case the vector space $V$ is finite dimensional it has a finite spanning set say $X={x_1,dots,x_n}$. The independent set $S$ is going to have cardinality atmost $n$(see Steinitz exchange theorem). Now keep adding vectors from $X$ to $S$ which are not there in the span of the vectors already there. For instance if $x_1$ is there in span($S$) ignore otherwise adjoin to get $Scup{x_1}$. Now if $x_2$ is there in span($Scup{x_1}$) or span($S$)(by whatever happened in previous step) then ignore otherwise adjoin to get $Scup{x_1,x_2}$ or $Scup{x_2}$. Continue like this for $n$ steps to get a linearly independent set (since none of the vectors are in the span of the previous ones) containing $S$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 13:41









                  Arpan DasArpan Das

                  837




                  837






























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