Isomorphism of X to itself. [closed]












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Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.










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closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    In this generality it's not true. Are there any connections between the operations?
    $endgroup$
    – Berci
    Dec 26 '18 at 12:55










  • $begingroup$
    Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
    $endgroup$
    – HelloDarkness
    Dec 26 '18 at 12:59








  • 2




    $begingroup$
    The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
    $endgroup$
    – Berci
    Dec 26 '18 at 13:04
















0












$begingroup$


Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.










share|cite|improve this question









$endgroup$



closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    In this generality it's not true. Are there any connections between the operations?
    $endgroup$
    – Berci
    Dec 26 '18 at 12:55










  • $begingroup$
    Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
    $endgroup$
    – HelloDarkness
    Dec 26 '18 at 12:59








  • 2




    $begingroup$
    The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
    $endgroup$
    – Berci
    Dec 26 '18 at 13:04














0












0








0


1



$begingroup$


Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.










share|cite|improve this question









$endgroup$




Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.







linear-algebra vector-space-isomorphism






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asked Dec 26 '18 at 12:49









HelloDarknessHelloDarkness

215




215




closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    In this generality it's not true. Are there any connections between the operations?
    $endgroup$
    – Berci
    Dec 26 '18 at 12:55










  • $begingroup$
    Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
    $endgroup$
    – HelloDarkness
    Dec 26 '18 at 12:59








  • 2




    $begingroup$
    The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
    $endgroup$
    – Berci
    Dec 26 '18 at 13:04














  • 1




    $begingroup$
    In this generality it's not true. Are there any connections between the operations?
    $endgroup$
    – Berci
    Dec 26 '18 at 12:55










  • $begingroup$
    Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
    $endgroup$
    – HelloDarkness
    Dec 26 '18 at 12:59








  • 2




    $begingroup$
    The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
    $endgroup$
    – Berci
    Dec 26 '18 at 13:04








1




1




$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55




$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55












$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59






$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59






2




2




$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04




$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04










2 Answers
2






active

oldest

votes


















1












$begingroup$

Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:



$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$



if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$






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$endgroup$





















    0












    $begingroup$

    The fact that the set is the same is a distraction.
    Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
    Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
    (Check that this defines a linear map)



    (This answer assumes that the dimensions are the same)






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:



      $$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$



      if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:



        $$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$



        if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:



          $$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$



          if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$






          share|cite|improve this answer









          $endgroup$



          Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:



          $$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$



          if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 13:04









          Math_QEDMath_QED

          7,64031452




          7,64031452























              0












              $begingroup$

              The fact that the set is the same is a distraction.
              Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
              Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
              (Check that this defines a linear map)



              (This answer assumes that the dimensions are the same)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The fact that the set is the same is a distraction.
                Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
                Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
                (Check that this defines a linear map)



                (This answer assumes that the dimensions are the same)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The fact that the set is the same is a distraction.
                  Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
                  Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
                  (Check that this defines a linear map)



                  (This answer assumes that the dimensions are the same)






                  share|cite|improve this answer









                  $endgroup$



                  The fact that the set is the same is a distraction.
                  Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
                  Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
                  (Check that this defines a linear map)



                  (This answer assumes that the dimensions are the same)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 13:01









                  GalGal

                  92




                  92















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