Equivalence of surjectivity and injectivity for linear operators on finite dimensional vector spaces












0












$begingroup$


I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by



Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$



Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.



$w = v^i(T e_i) = v^ie'_i$.



Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.



Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.



Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?






linear-transformations







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$endgroup$












  • $begingroup$
    Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:26










  • $begingroup$
    For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:27










  • $begingroup$
    Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
    $endgroup$
    – user2582713
    Jul 16 '16 at 13:12










  • $begingroup$
    I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
    $endgroup$
    – John Coleman
    Jul 16 '16 at 13:15










  • $begingroup$
    @John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
    $endgroup$
    – user39082
    Jul 16 '16 at 13:18
















0












$begingroup$


I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by



Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$



Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.



$w = v^i(T e_i) = v^ie'_i$.



Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.



Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.



Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:26










  • $begingroup$
    For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:27










  • $begingroup$
    Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
    $endgroup$
    – user2582713
    Jul 16 '16 at 13:12










  • $begingroup$
    I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
    $endgroup$
    – John Coleman
    Jul 16 '16 at 13:15










  • $begingroup$
    @John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
    $endgroup$
    – user39082
    Jul 16 '16 at 13:18














0












0








0





$begingroup$


I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by



Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$



Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.



$w = v^i(T e_i) = v^ie'_i$.



Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.



Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.



Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?






linear-transformations







share|cite|improve this question











$endgroup$




I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by



Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$



Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.



$w = v^i(T e_i) = v^ie'_i$.



Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.



Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.



Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?






linear-transformations




linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 16 '16 at 13:02







user2582713

















asked Jul 16 '16 at 11:36









user2582713user2582713

136




136












  • $begingroup$
    Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:26










  • $begingroup$
    For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:27










  • $begingroup$
    Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
    $endgroup$
    – user2582713
    Jul 16 '16 at 13:12










  • $begingroup$
    I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
    $endgroup$
    – John Coleman
    Jul 16 '16 at 13:15










  • $begingroup$
    @John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
    $endgroup$
    – user39082
    Jul 16 '16 at 13:18


















  • $begingroup$
    Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:26










  • $begingroup$
    For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
    $endgroup$
    – user39082
    Jul 16 '16 at 12:27










  • $begingroup$
    Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
    $endgroup$
    – user2582713
    Jul 16 '16 at 13:12










  • $begingroup$
    I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
    $endgroup$
    – John Coleman
    Jul 16 '16 at 13:15










  • $begingroup$
    @John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
    $endgroup$
    – user39082
    Jul 16 '16 at 13:18
















$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26




$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26












$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27




$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27












$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12




$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12












$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15




$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15












$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18




$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18










2 Answers
2






active

oldest

votes


















0












$begingroup$

No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.



However the equivalence is true for finite-dimensional vector spaces.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
    $endgroup$
    – user2582713
    Jul 16 '16 at 11:58



















0












$begingroup$

Surjectivity ===> Injectivity:



Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.



Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.



But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.



Injectivity ===> Surjectivity



The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.






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









    0












    $begingroup$

    No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
    $$Te_i=e_{i+1} forall iin{mathbb N}$$
    defines an injective but not surjective operator, and
    $$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
    defines a surjective but not injective operator.



    However the equivalence is true for finite-dimensional vector spaces.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
      $endgroup$
      – user2582713
      Jul 16 '16 at 11:58
















    0












    $begingroup$

    No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
    $$Te_i=e_{i+1} forall iin{mathbb N}$$
    defines an injective but not surjective operator, and
    $$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
    defines a surjective but not injective operator.



    However the equivalence is true for finite-dimensional vector spaces.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
      $endgroup$
      – user2582713
      Jul 16 '16 at 11:58














    0












    0








    0





    $begingroup$

    No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
    $$Te_i=e_{i+1} forall iin{mathbb N}$$
    defines an injective but not surjective operator, and
    $$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
    defines a surjective but not injective operator.



    However the equivalence is true for finite-dimensional vector spaces.






    share|cite|improve this answer









    $endgroup$



    No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
    $$Te_i=e_{i+1} forall iin{mathbb N}$$
    defines an injective but not surjective operator, and
    $$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
    defines a surjective but not injective operator.



    However the equivalence is true for finite-dimensional vector spaces.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 16 '16 at 11:49









    user39082user39082

    1,237513




    1,237513












    • $begingroup$
      Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
      $endgroup$
      – user2582713
      Jul 16 '16 at 11:58


















    • $begingroup$
      Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
      $endgroup$
      – user2582713
      Jul 16 '16 at 11:58
















    $begingroup$
    Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
    $endgroup$
    – user2582713
    Jul 16 '16 at 11:58




    $begingroup$
    Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
    $endgroup$
    – user2582713
    Jul 16 '16 at 11:58











    0












    $begingroup$

    Surjectivity ===> Injectivity:



    Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.



    Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.



    But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.



    Injectivity ===> Surjectivity



    The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Surjectivity ===> Injectivity:



      Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.



      Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.



      But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.



      Injectivity ===> Surjectivity



      The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Surjectivity ===> Injectivity:



        Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.



        Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.



        But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.



        Injectivity ===> Surjectivity



        The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.






        share|cite|improve this answer









        $endgroup$



        Surjectivity ===> Injectivity:



        Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.



        Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.



        But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.



        Injectivity ===> Surjectivity



        The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 16 '16 at 12:23









        user39082user39082

        1,237513




        1,237513






























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