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







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    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






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      asked Jan 3 at 22:54









      JOHN

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          2 Answers
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          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





















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              2 Answers
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              2 Answers
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              active

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              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












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                          share|cite|improve this answer










                          answered Jan 3 at 23:05









                          Foobaz John

                          21.4k41351




                          21.4k41351






























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