Prove span is the smallest containing subspace












0














In Linear Algebra Done Right, it proved that the span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.



I followed the proof that $span(v_1,...,v_m)$ is a subspace of $V$. But I don't follow the proof of smallest subspace.



Each $v_j$ is a linear combination of $v_1,...,v_m$ (to show this, set $a_j = 1$ and let the other $a$'s in $a_1v_1+...+a_mv_m$ equal $0$) [I don't understand the sentence in the parenthesis. What is the message?]. Thus $span(v_1,...v_m)$ contains each $v_j$.



Conversely, because subspaces are closed under scalar multiplication and addition, every subspace of $V$ containing each $v_j$ contains $span(v_1,...v_m)$ [I don't understand this line].



Thus $span(v_1,...v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,...v_m$ [how to reach this conclusion?]






linear-algebra







share|cite|improve this question



























    0














    In Linear Algebra Done Right, it proved that the span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.



    I followed the proof that $span(v_1,...,v_m)$ is a subspace of $V$. But I don't follow the proof of smallest subspace.



    Each $v_j$ is a linear combination of $v_1,...,v_m$ (to show this, set $a_j = 1$ and let the other $a$'s in $a_1v_1+...+a_mv_m$ equal $0$) [I don't understand the sentence in the parenthesis. What is the message?]. Thus $span(v_1,...v_m)$ contains each $v_j$.



    Conversely, because subspaces are closed under scalar multiplication and addition, every subspace of $V$ containing each $v_j$ contains $span(v_1,...v_m)$ [I don't understand this line].



    Thus $span(v_1,...v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,...v_m$ [how to reach this conclusion?]






    linear-algebra







    share|cite|improve this question

























      0












      0








      0







      In Linear Algebra Done Right, it proved that the span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.



      I followed the proof that $span(v_1,...,v_m)$ is a subspace of $V$. But I don't follow the proof of smallest subspace.



      Each $v_j$ is a linear combination of $v_1,...,v_m$ (to show this, set $a_j = 1$ and let the other $a$'s in $a_1v_1+...+a_mv_m$ equal $0$) [I don't understand the sentence in the parenthesis. What is the message?]. Thus $span(v_1,...v_m)$ contains each $v_j$.



      Conversely, because subspaces are closed under scalar multiplication and addition, every subspace of $V$ containing each $v_j$ contains $span(v_1,...v_m)$ [I don't understand this line].



      Thus $span(v_1,...v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,...v_m$ [how to reach this conclusion?]






      linear-algebra







      share|cite|improve this question













      In Linear Algebra Done Right, it proved that the span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.



      I followed the proof that $span(v_1,...,v_m)$ is a subspace of $V$. But I don't follow the proof of smallest subspace.



      Each $v_j$ is a linear combination of $v_1,...,v_m$ (to show this, set $a_j = 1$ and let the other $a$'s in $a_1v_1+...+a_mv_m$ equal $0$) [I don't understand the sentence in the parenthesis. What is the message?]. Thus $span(v_1,...v_m)$ contains each $v_j$.



      Conversely, because subspaces are closed under scalar multiplication and addition, every subspace of $V$ containing each $v_j$ contains $span(v_1,...v_m)$ [I don't understand this line].



      Thus $span(v_1,...v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,...v_m$ [how to reach this conclusion?]






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 22:54









      JOHN

      826




      826






















          2 Answers
          2






          active

          oldest

          votes


















          1














          1) We wish to express $v_j$ as a linear combination of the vectors $v_1, dots, v_m$. So suppose
          $$v_j = a_1v_1 + dots + a_mv_m.$$
          Then taking $a_j = 1$ and $a_i = 0$ for $i neq j$ does the trick.



          2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, dots, v_m$. That is, it contains their span.



          3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace.






          share|cite|improve this answer





























            0














            Let $M$ be the smallest vector subspace of $V$ containing $v_1,dotsc, v_n$. We claim that $M=text{span}(v_1,dotsc,v_n)$. Indeed, note that
            $$
            v_i=0cdot v_1+0cdot v_2+dotsb+1cdot v_i+dotsb+0cdot v_n
            $$
            for all $1leq ileq n$. So by the definition of the span as the collection of all linear combinations of $v_1,dotsc, v_n$, we have that
            $$
            v_iintext{span}(v_1,dotsc, v_n)
            $$
            for all $1leq ileq n$. But $M$ is the smallest vector subspace containing $v_1,dots, v_n$, so
            $$
            Msubseteq text{span}(v_1,dotsc, v_n).
            $$
            For the reverse inclusion, since $v_iin M$ for $1leq ileq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
            $$
            a_1v_1+dotsb+a_nv_nin M
            $$
            for all $(a_1,dotsc, a_n)inmathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,dotsc, v_n$). So
            $$
            text{span}(v_1,dotsc, v_n)subseteq M.
            $$






            share|cite|improve this answer





















              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%2f3061117%2fprove-span-is-the-smallest-containing-subspace%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              1) We wish to express $v_j$ as a linear combination of the vectors $v_1, dots, v_m$. So suppose
              $$v_j = a_1v_1 + dots + a_mv_m.$$
              Then taking $a_j = 1$ and $a_i = 0$ for $i neq j$ does the trick.



              2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, dots, v_m$. That is, it contains their span.



              3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace.






              share|cite|improve this answer


























                1














                1) We wish to express $v_j$ as a linear combination of the vectors $v_1, dots, v_m$. So suppose
                $$v_j = a_1v_1 + dots + a_mv_m.$$
                Then taking $a_j = 1$ and $a_i = 0$ for $i neq j$ does the trick.



                2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, dots, v_m$. That is, it contains their span.



                3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace.






                share|cite|improve this answer
























                  1












                  1








                  1






                  1) We wish to express $v_j$ as a linear combination of the vectors $v_1, dots, v_m$. So suppose
                  $$v_j = a_1v_1 + dots + a_mv_m.$$
                  Then taking $a_j = 1$ and $a_i = 0$ for $i neq j$ does the trick.



                  2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, dots, v_m$. That is, it contains their span.



                  3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace.






                  share|cite|improve this answer












                  1) We wish to express $v_j$ as a linear combination of the vectors $v_1, dots, v_m$. So suppose
                  $$v_j = a_1v_1 + dots + a_mv_m.$$
                  Then taking $a_j = 1$ and $a_i = 0$ for $i neq j$ does the trick.



                  2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, dots, v_m$. That is, it contains their span.



                  3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 23:04









                  bounceback

                  1846




                  1846























                      0














                      Let $M$ be the smallest vector subspace of $V$ containing $v_1,dotsc, v_n$. We claim that $M=text{span}(v_1,dotsc,v_n)$. Indeed, note that
                      $$
                      v_i=0cdot v_1+0cdot v_2+dotsb+1cdot v_i+dotsb+0cdot v_n
                      $$
                      for all $1leq ileq n$. So by the definition of the span as the collection of all linear combinations of $v_1,dotsc, v_n$, we have that
                      $$
                      v_iintext{span}(v_1,dotsc, v_n)
                      $$
                      for all $1leq ileq n$. But $M$ is the smallest vector subspace containing $v_1,dots, v_n$, so
                      $$
                      Msubseteq text{span}(v_1,dotsc, v_n).
                      $$
                      For the reverse inclusion, since $v_iin M$ for $1leq ileq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
                      $$
                      a_1v_1+dotsb+a_nv_nin M
                      $$
                      for all $(a_1,dotsc, a_n)inmathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,dotsc, v_n$). So
                      $$
                      text{span}(v_1,dotsc, v_n)subseteq M.
                      $$






                      share|cite|improve this answer


























                        0














                        Let $M$ be the smallest vector subspace of $V$ containing $v_1,dotsc, v_n$. We claim that $M=text{span}(v_1,dotsc,v_n)$. Indeed, note that
                        $$
                        v_i=0cdot v_1+0cdot v_2+dotsb+1cdot v_i+dotsb+0cdot v_n
                        $$
                        for all $1leq ileq n$. So by the definition of the span as the collection of all linear combinations of $v_1,dotsc, v_n$, we have that
                        $$
                        v_iintext{span}(v_1,dotsc, v_n)
                        $$
                        for all $1leq ileq n$. But $M$ is the smallest vector subspace containing $v_1,dots, v_n$, so
                        $$
                        Msubseteq text{span}(v_1,dotsc, v_n).
                        $$
                        For the reverse inclusion, since $v_iin M$ for $1leq ileq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
                        $$
                        a_1v_1+dotsb+a_nv_nin M
                        $$
                        for all $(a_1,dotsc, a_n)inmathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,dotsc, v_n$). So
                        $$
                        text{span}(v_1,dotsc, v_n)subseteq M.
                        $$






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Let $M$ be the smallest vector subspace of $V$ containing $v_1,dotsc, v_n$. We claim that $M=text{span}(v_1,dotsc,v_n)$. Indeed, note that
                          $$
                          v_i=0cdot v_1+0cdot v_2+dotsb+1cdot v_i+dotsb+0cdot v_n
                          $$
                          for all $1leq ileq n$. So by the definition of the span as the collection of all linear combinations of $v_1,dotsc, v_n$, we have that
                          $$
                          v_iintext{span}(v_1,dotsc, v_n)
                          $$
                          for all $1leq ileq n$. But $M$ is the smallest vector subspace containing $v_1,dots, v_n$, so
                          $$
                          Msubseteq text{span}(v_1,dotsc, v_n).
                          $$
                          For the reverse inclusion, since $v_iin M$ for $1leq ileq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
                          $$
                          a_1v_1+dotsb+a_nv_nin M
                          $$
                          for all $(a_1,dotsc, a_n)inmathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,dotsc, v_n$). So
                          $$
                          text{span}(v_1,dotsc, v_n)subseteq M.
                          $$






                          share|cite|improve this answer












                          Let $M$ be the smallest vector subspace of $V$ containing $v_1,dotsc, v_n$. We claim that $M=text{span}(v_1,dotsc,v_n)$. Indeed, note that
                          $$
                          v_i=0cdot v_1+0cdot v_2+dotsb+1cdot v_i+dotsb+0cdot v_n
                          $$
                          for all $1leq ileq n$. So by the definition of the span as the collection of all linear combinations of $v_1,dotsc, v_n$, we have that
                          $$
                          v_iintext{span}(v_1,dotsc, v_n)
                          $$
                          for all $1leq ileq n$. But $M$ is the smallest vector subspace containing $v_1,dots, v_n$, so
                          $$
                          Msubseteq text{span}(v_1,dotsc, v_n).
                          $$
                          For the reverse inclusion, since $v_iin M$ for $1leq ileq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
                          $$
                          a_1v_1+dotsb+a_nv_nin M
                          $$
                          for all $(a_1,dotsc, a_n)inmathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,dotsc, v_n$). So
                          $$
                          text{span}(v_1,dotsc, v_n)subseteq M.
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 3 at 23:05









                          Foobaz John

                          21.4k41351




                          21.4k41351






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3061117%2fprove-span-is-the-smallest-containing-subspace%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

                              1300-talet

                              Image
                              Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                              Read more

                              1300-talet

                              Image
                              Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                              Read more

                              Display a custom attribute below product name in the front-end Magento 1.9.3.8

                              Image
                              0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
                              Read more