Why is the Change of Basis map unique?












5












$begingroup$


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 $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give 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 me as to why this is necessarily true?






linear-algebra linear-transformations







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$endgroup$








  • 5




    $begingroup$
    The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
    $endgroup$
    – M. Winter
    Feb 7 at 9:22












  • $begingroup$
    Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
    $endgroup$
    – Evpok
    Feb 8 at 12:42


















5












$begingroup$


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 $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give 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 me as to why this is necessarily true?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
    $endgroup$
    – M. Winter
    Feb 7 at 9:22












  • $begingroup$
    Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
    $endgroup$
    – Evpok
    Feb 8 at 12:42
















5












5








5


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 $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give 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 me 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 $mathcal{C}={u_1,...,u_n}$, the coordinate maps $_{mathcal{B}}:Vrightarrow mathbb{F}^n$ and $_{mathcal{C}}:Vrightarrow mathbb{F}^n$ give 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 me as to why this is necessarily true?






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 8 at 1:19









J. W. Tanner

4,0611320




4,0611320










asked Feb 7 at 6:50









Joe Man AnalysisJoe Man Analysis

56619




56619








  • 5




    $begingroup$
    The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
    $endgroup$
    – M. Winter
    Feb 7 at 9:22












  • $begingroup$
    Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
    $endgroup$
    – Evpok
    Feb 8 at 12:42
















  • 5




    $begingroup$
    The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
    $endgroup$
    – M. Winter
    Feb 7 at 9:22












  • $begingroup$
    Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
    $endgroup$
    – Evpok
    Feb 8 at 12:42










5




5




$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22






$begingroup$
The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $coscircsin^{-1}$ unique?".
$endgroup$
– M. Winter
Feb 7 at 9:22














$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42






$begingroup$
Because if it was not unique, there would be at least a vector with several possible coordinates in $mathcal C$
$endgroup$
– Evpok
Feb 8 at 12:42












2 Answers
2






active

oldest

votes


















14












$begingroup$

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









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      14












      $begingroup$

      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









      $endgroup$


















        14












        $begingroup$

        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









        $endgroup$
















          14












          14








          14





          $begingroup$

          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 Feb 7 at 6:56









          DonAntonioDonAntonio

          180k1494233




          180k1494233























              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 Feb 7 at 6:57









                  AlessioDVAlessioDV

                  1,158114




                  1,158114






























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