If $n$ vectors are linearly independent, is their span $mathbb{R}^n$?












3












$begingroup$


Have $n$ vectors in $mathbb{R}^n$.



If the $n$ vectors are linearly independent, can we conclude that their span is $mathbb{R}^n$?






linear-algebra geometry vectors







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Yes${}{}{}{}{}{}$.
    $endgroup$
    – user63181
    May 26 '14 at 23:00










  • $begingroup$
    How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
    $endgroup$
    – Muphrid
    May 26 '14 at 23:01










  • $begingroup$
    @Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
    $endgroup$
    – Zol Tun Kul
    May 26 '14 at 23:02
















3












$begingroup$


Have $n$ vectors in $mathbb{R}^n$.



If the $n$ vectors are linearly independent, can we conclude that their span is $mathbb{R}^n$?






linear-algebra geometry vectors







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Yes${}{}{}{}{}{}$.
    $endgroup$
    – user63181
    May 26 '14 at 23:00










  • $begingroup$
    How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
    $endgroup$
    – Muphrid
    May 26 '14 at 23:01










  • $begingroup$
    @Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
    $endgroup$
    – Zol Tun Kul
    May 26 '14 at 23:02














3












3








3


2



$begingroup$


Have $n$ vectors in $mathbb{R}^n$.



If the $n$ vectors are linearly independent, can we conclude that their span is $mathbb{R}^n$?






linear-algebra geometry vectors







share|cite|improve this question











$endgroup$




Have $n$ vectors in $mathbb{R}^n$.



If the $n$ vectors are linearly independent, can we conclude that their span is $mathbb{R}^n$?






linear-algebra geometry vectors




linear-algebra geometry vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 26 '14 at 23:01









Git Gud

28.9k1050101




28.9k1050101










asked May 26 '14 at 23:00









Zol Tun KulZol Tun Kul

2,84682951




2,84682951








  • 5




    $begingroup$
    Yes${}{}{}{}{}{}$.
    $endgroup$
    – user63181
    May 26 '14 at 23:00










  • $begingroup$
    How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
    $endgroup$
    – Muphrid
    May 26 '14 at 23:01










  • $begingroup$
    @Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
    $endgroup$
    – Zol Tun Kul
    May 26 '14 at 23:02














  • 5




    $begingroup$
    Yes${}{}{}{}{}{}$.
    $endgroup$
    – user63181
    May 26 '14 at 23:00










  • $begingroup$
    How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
    $endgroup$
    – Muphrid
    May 26 '14 at 23:01










  • $begingroup$
    @Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
    $endgroup$
    – Zol Tun Kul
    May 26 '14 at 23:02








5




5




$begingroup$
Yes${}{}{}{}{}{}$.
$endgroup$
– user63181
May 26 '14 at 23:00




$begingroup$
Yes${}{}{}{}{}{}$.
$endgroup$
– user63181
May 26 '14 at 23:00












$begingroup$
How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
$endgroup$
– Muphrid
May 26 '14 at 23:01




$begingroup$
How do you define span? Is this a conclusion that might follow from a definition you're familiar with?
$endgroup$
– Muphrid
May 26 '14 at 23:01












$begingroup$
@Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
$endgroup$
– Zol Tun Kul
May 26 '14 at 23:02




$begingroup$
@Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors.
$endgroup$
– Zol Tun Kul
May 26 '14 at 23:02










3 Answers
3






active

oldest

votes


















0












$begingroup$

A useful result in linear algebra is:



Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $dim W=dim V$, then $W=V$.



If we are given linearly independent $v_1,dotsc,v_ninBbb R^n$, then $DeclareMathOperator{Span}{Span}Span{v_1,dotsc,v_n}$ is an $n$-dimensional subspace of $Bbb R^n$. Since $dim Bbb R^n=n$, the proposition implies that $Span{v_1,dotsc,v_n}=Bbb R^n$.



Can you prove this proposition?






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $mathbb{R}^n$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
      $endgroup$
      – Thomas Andrews
      May 26 '14 at 23:22










    • $begingroup$
      True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
      $endgroup$
      – Michael M
      May 27 '14 at 1:06



















    0












    $begingroup$

    Hint: if the span of $m$ linearly independent vectors ${v_1,dots,v_m}$ is a proper subspace $U$ of $mathbb{R}^n$, then, for every vector $vinmathbb{R}$, $vnotin U$, the set ${v_1,dots,v_m,v}$ is linearly independent.



    Can you see there's a contradiction if $m=n$?






    share|cite|improve this answer









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f810551%2fif-n-vectors-are-linearly-independent-is-their-span-mathbbrn%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      A useful result in linear algebra is:



      Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $dim W=dim V$, then $W=V$.



      If we are given linearly independent $v_1,dotsc,v_ninBbb R^n$, then $DeclareMathOperator{Span}{Span}Span{v_1,dotsc,v_n}$ is an $n$-dimensional subspace of $Bbb R^n$. Since $dim Bbb R^n=n$, the proposition implies that $Span{v_1,dotsc,v_n}=Bbb R^n$.



      Can you prove this proposition?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        A useful result in linear algebra is:



        Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $dim W=dim V$, then $W=V$.



        If we are given linearly independent $v_1,dotsc,v_ninBbb R^n$, then $DeclareMathOperator{Span}{Span}Span{v_1,dotsc,v_n}$ is an $n$-dimensional subspace of $Bbb R^n$. Since $dim Bbb R^n=n$, the proposition implies that $Span{v_1,dotsc,v_n}=Bbb R^n$.



        Can you prove this proposition?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          A useful result in linear algebra is:



          Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $dim W=dim V$, then $W=V$.



          If we are given linearly independent $v_1,dotsc,v_ninBbb R^n$, then $DeclareMathOperator{Span}{Span}Span{v_1,dotsc,v_n}$ is an $n$-dimensional subspace of $Bbb R^n$. Since $dim Bbb R^n=n$, the proposition implies that $Span{v_1,dotsc,v_n}=Bbb R^n$.



          Can you prove this proposition?






          share|cite|improve this answer









          $endgroup$



          A useful result in linear algebra is:



          Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $dim W=dim V$, then $W=V$.



          If we are given linearly independent $v_1,dotsc,v_ninBbb R^n$, then $DeclareMathOperator{Span}{Span}Span{v_1,dotsc,v_n}$ is an $n$-dimensional subspace of $Bbb R^n$. Since $dim Bbb R^n=n$, the proposition implies that $Span{v_1,dotsc,v_n}=Bbb R^n$.



          Can you prove this proposition?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 26 '14 at 23:08









          Brian FitzpatrickBrian Fitzpatrick

          21.8k42959




          21.8k42959























              2












              $begingroup$

              You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $mathbb{R}^n$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
                $endgroup$
                – Thomas Andrews
                May 26 '14 at 23:22










              • $begingroup$
                True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
                $endgroup$
                – Michael M
                May 27 '14 at 1:06
















              2












              $begingroup$

              You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $mathbb{R}^n$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
                $endgroup$
                – Thomas Andrews
                May 26 '14 at 23:22










              • $begingroup$
                True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
                $endgroup$
                – Michael M
                May 27 '14 at 1:06














              2












              2








              2





              $begingroup$

              You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $mathbb{R}^n$.






              share|cite|improve this answer









              $endgroup$



              You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $mathbb{R}^n$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered May 26 '14 at 23:10









              Michael MMichael M

              2,183614




              2,183614












              • $begingroup$
                That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
                $endgroup$
                – Thomas Andrews
                May 26 '14 at 23:22










              • $begingroup$
                True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
                $endgroup$
                – Michael M
                May 27 '14 at 1:06


















              • $begingroup$
                That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
                $endgroup$
                – Thomas Andrews
                May 26 '14 at 23:22










              • $begingroup$
                True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
                $endgroup$
                – Michael M
                May 27 '14 at 1:06
















              $begingroup$
              That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
              $endgroup$
              – Thomas Andrews
              May 26 '14 at 23:22




              $begingroup$
              That assumes some theorems about dimension, of course - namely, that dimension is well-defined...
              $endgroup$
              – Thomas Andrews
              May 26 '14 at 23:22












              $begingroup$
              True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
              $endgroup$
              – Michael M
              May 27 '14 at 1:06




              $begingroup$
              True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal.
              $endgroup$
              – Michael M
              May 27 '14 at 1:06











              0












              $begingroup$

              Hint: if the span of $m$ linearly independent vectors ${v_1,dots,v_m}$ is a proper subspace $U$ of $mathbb{R}^n$, then, for every vector $vinmathbb{R}$, $vnotin U$, the set ${v_1,dots,v_m,v}$ is linearly independent.



              Can you see there's a contradiction if $m=n$?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint: if the span of $m$ linearly independent vectors ${v_1,dots,v_m}$ is a proper subspace $U$ of $mathbb{R}^n$, then, for every vector $vinmathbb{R}$, $vnotin U$, the set ${v_1,dots,v_m,v}$ is linearly independent.



                Can you see there's a contradiction if $m=n$?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint: if the span of $m$ linearly independent vectors ${v_1,dots,v_m}$ is a proper subspace $U$ of $mathbb{R}^n$, then, for every vector $vinmathbb{R}$, $vnotin U$, the set ${v_1,dots,v_m,v}$ is linearly independent.



                  Can you see there's a contradiction if $m=n$?






                  share|cite|improve this answer









                  $endgroup$



                  Hint: if the span of $m$ linearly independent vectors ${v_1,dots,v_m}$ is a proper subspace $U$ of $mathbb{R}^n$, then, for every vector $vinmathbb{R}$, $vnotin U$, the set ${v_1,dots,v_m,v}$ is linearly independent.



                  Can you see there's a contradiction if $m=n$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 26 '14 at 23:11









                  egregegreg

                  185k1486206




                  185k1486206






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f810551%2fif-n-vectors-are-linearly-independent-is-their-span-mathbbrn%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Brian Clough

                      Image
                      Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
                      Read more

                      Cáceres

                      Image
                      Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
                      Read more