Why is the Change of Basis map unique?












4












$begingroup$


I've been looking all over, but i haven't found anything satisfactory.



We've been shown in class by a commutative diagram that given an $n$-dimensional vector space $V$ over a field, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, that the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ gives rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$ which is our change of basis matrix.



But I am having a lot of trouble proving that P is unique. Can anyone enlighten my as to why this is necessarily true?






linear-algebra linear-transformations







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$endgroup$

















    4












    $begingroup$


    I've been looking all over, but i haven't found anything satisfactory.



    We've been shown in class by a commutative diagram that given an $n$-dimensional vector space $V$ over a field, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, that the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ gives rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$ which is our change of basis matrix.



    But I am having a lot of trouble proving that P is unique. Can anyone enlighten my as to why this is necessarily true?






    linear-algebra linear-transformations







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I've been looking all over, but i haven't found anything satisfactory.



      We've been shown in class by a commutative diagram that given an $n$-dimensional vector space $V$ over a field, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, that the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ gives rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$ which is our change of basis matrix.



      But I am having a lot of trouble proving that P is unique. Can anyone enlighten my as to why this is necessarily true?






      linear-algebra linear-transformations







      share|cite|improve this question









      $endgroup$




      I've been looking all over, but i haven't found anything satisfactory.



      We've been shown in class by a commutative diagram that given an $n$-dimensional vector space $V$ over a field, $mathbb{F}$, and bases, $mathcal{B}={v_1,...,v_n}$ and $mathcal{C}={u_1,...,u_n}$, that the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ gives rise to a unique map $P=_{mathcal{B}}circ ^{-1}_{mathcal{C}}:mathbb{F}^nrightarrow mathbb{F}^n$ which is our change of basis matrix.



      But I am having a lot of trouble proving that P is unique. Can anyone enlighten my as to why this is necessarily true?






      linear-algebra linear-transformations




      linear-algebra linear-transformations






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      share|cite|improve this question











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      asked 58 mins ago









      Joe Man AnalysisJoe Man Analysis

      36419




      36419






















          2 Answers
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          3












          $begingroup$

          Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.



          If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any



          $$v=sum_{k=1}^n a_iv_iin V;,;;Qv=sum_{k=1}^na_iQv_i=sum_{k=1}^n a_iPv_i=Pv$$



          so $;Qequiv P;$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.






            share|cite|improve this answer









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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.



              If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any



              $$v=sum_{k=1}^n a_iv_iin V;,;;Qv=sum_{k=1}^na_iQv_i=sum_{k=1}^n a_iPv_i=Pv$$



              so $;Qequiv P;$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.



                If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any



                $$v=sum_{k=1}^n a_iv_iin V;,;;Qv=sum_{k=1}^na_iQv_i=sum_{k=1}^n a_iPv_i=Pv$$



                so $;Qequiv P;$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.



                  If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any



                  $$v=sum_{k=1}^n a_iv_iin V;,;;Qv=sum_{k=1}^na_iQv_i=sum_{k=1}^n a_iPv_i=Pv$$



                  so $;Qequiv P;$.






                  share|cite|improve this answer









                  $endgroup$



                  Remember that any linear map on any linear space $;V;$ is uniquely and completely determined once we know its action on any basis of $;V;$ ...and that's all.



                  If you want to do this proof, suppose there's another map $;Q:Vto V;$ s.t. it coincides on "the old basis" $;mathcal B;$ with $;P:;; Qv_i=Pv_i;;forall,i=1,2,...,n;$ , then (using linearity of the maps), for any



                  $$v=sum_{k=1}^n a_iv_iin V;,;;Qv=sum_{k=1}^na_iQv_i=sum_{k=1}^n a_iPv_i=Pv$$



                  so $;Qequiv P;$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 53 mins ago









                  DonAntonioDonAntonio

                  178k1493229




                  178k1493229























                      2












                      $begingroup$

                      A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.






                          share|cite|improve this answer









                          $endgroup$



                          A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $mathcal B$ and $mathcal C$ as in your post, then a change of basis $f$ from $mathcal B$ to $mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $mathcal B$ then there is only a unique map that satisfies those conditions.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 51 mins ago









                          AlessioDVAlessioDV

                          2287




                          2287






























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