If ${ u_1,…,u_n}$ is a linearly independent of $U$, show that ${ T(u_1), …, T(u_n)}$ is linearly...












1












$begingroup$


My Professor gives us a homework that contains this question:




Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U to V$ be a linear transformation on $F$.



If ${ u_1,...,u_n}$ is a linearly independent of $U$, show that ${ T(u_1), ..., T(u_n)}$ is linearly independent.




Is that true? I took $T:mathbb{R}^2 to mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) in mathbb{R}^2$



Then I found that ${ (1,0),(0,1)}$ is linearly independent in $ mathbb{R}^2$, but ${T(1,0)=0,T(0,1)=0}$ is not linearly independent in $mathbb{R}$ ?



What do you think?






linear-algebra linear-transformations







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$endgroup$








  • 1




    $begingroup$
    I think you are correct, and you may need to check if it is linearly dependent or independent.
    $endgroup$
    – Quang Hoang
    Nov 28 '18 at 17:19






  • 1




    $begingroup$
    It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:21






  • 1




    $begingroup$
    Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:29
















1












$begingroup$


My Professor gives us a homework that contains this question:




Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U to V$ be a linear transformation on $F$.



If ${ u_1,...,u_n}$ is a linearly independent of $U$, show that ${ T(u_1), ..., T(u_n)}$ is linearly independent.




Is that true? I took $T:mathbb{R}^2 to mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) in mathbb{R}^2$



Then I found that ${ (1,0),(0,1)}$ is linearly independent in $ mathbb{R}^2$, but ${T(1,0)=0,T(0,1)=0}$ is not linearly independent in $mathbb{R}$ ?



What do you think?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think you are correct, and you may need to check if it is linearly dependent or independent.
    $endgroup$
    – Quang Hoang
    Nov 28 '18 at 17:19






  • 1




    $begingroup$
    It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:21






  • 1




    $begingroup$
    Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:29














1












1








1





$begingroup$


My Professor gives us a homework that contains this question:




Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U to V$ be a linear transformation on $F$.



If ${ u_1,...,u_n}$ is a linearly independent of $U$, show that ${ T(u_1), ..., T(u_n)}$ is linearly independent.




Is that true? I took $T:mathbb{R}^2 to mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) in mathbb{R}^2$



Then I found that ${ (1,0),(0,1)}$ is linearly independent in $ mathbb{R}^2$, but ${T(1,0)=0,T(0,1)=0}$ is not linearly independent in $mathbb{R}$ ?



What do you think?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$




My Professor gives us a homework that contains this question:




Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U to V$ be a linear transformation on $F$.



If ${ u_1,...,u_n}$ is a linearly independent of $U$, show that ${ T(u_1), ..., T(u_n)}$ is linearly independent.




Is that true? I took $T:mathbb{R}^2 to mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) in mathbb{R}^2$



Then I found that ${ (1,0),(0,1)}$ is linearly independent in $ mathbb{R}^2$, but ${T(1,0)=0,T(0,1)=0}$ is not linearly independent in $mathbb{R}$ ?



What do you think?






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 Nov 28 '18 at 17:17







Dima

















asked Nov 28 '18 at 17:14









DimaDima

623416




623416








  • 1




    $begingroup$
    I think you are correct, and you may need to check if it is linearly dependent or independent.
    $endgroup$
    – Quang Hoang
    Nov 28 '18 at 17:19






  • 1




    $begingroup$
    It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:21






  • 1




    $begingroup$
    Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:29














  • 1




    $begingroup$
    I think you are correct, and you may need to check if it is linearly dependent or independent.
    $endgroup$
    – Quang Hoang
    Nov 28 '18 at 17:19






  • 1




    $begingroup$
    It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:21






  • 1




    $begingroup$
    Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
    $endgroup$
    – Thomas Andrews
    Nov 28 '18 at 17:29








1




1




$begingroup$
I think you are correct, and you may need to check if it is linearly dependent or independent.
$endgroup$
– Quang Hoang
Nov 28 '18 at 17:19




$begingroup$
I think you are correct, and you may need to check if it is linearly dependent or independent.
$endgroup$
– Quang Hoang
Nov 28 '18 at 17:19




1




1




$begingroup$
It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
$endgroup$
– Thomas Andrews
Nov 28 '18 at 17:21




$begingroup$
It is not true. Indeed, you need $T$ to be an injection for this to be true in general.
$endgroup$
– Thomas Andrews
Nov 28 '18 at 17:21




1




1




$begingroup$
Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
$endgroup$
– Thomas Andrews
Nov 28 '18 at 17:29




$begingroup$
Or, if $T(u_1),dots,T(u_n)$ are independent, then $u_1,dots,u_n$ are independent.
$endgroup$
– Thomas Andrews
Nov 28 '18 at 17:29










3 Answers
3






active

oldest

votes


















6












$begingroup$

Your example is a valid counterexample to the claim.
So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),dots,T(u_n)$ are linearly independent, then $u_1,dots,u_n$ are linearly independent.




Additionally, a small note on terminology: it is preferable to say that ${ u_1,dots,u_n }$ is a linearly independent subset of $U$, and that ${ T(1,0), T(0,1) }$ is not linearly independent over $mathbb{R}$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
    $$alpha_1 T(u_1) + cdots + alpha_n T(u_n) = 0$$
    linearity tells us
    $$T(alpha_1 u_1 + cdots + alpha_n u_n) = 0,$$
    and hence if we apply $T^{-1}$ to both sides
    $$alpha_1 u_1 + cdots + alpha_n u_n = 0.$$ Since ${u_1,ldots,u_n}$ are linearly independent this implies that $alpha_1=cdots=alpha_n=0$, and then by definition ${T(u_1),ldots,T(u_n)}$ are linearly independent as well.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
      $endgroup$
      – Thomas Andrews
      Nov 28 '18 at 17:27



















    3












    $begingroup$

    In general, this isn't true as your counterexample shows.



    If $T$ is invertible then the statement holds. Start from
    $$c_1T(u_1)+cdots+c_nT(u_n)=0$$
    Apply $T^{-1}$ on both sides and get
    $$c_1u_1+cdots+c_nu_n=0$$
    and $c_1=cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      6












      $begingroup$

      Your example is a valid counterexample to the claim.
      So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),dots,T(u_n)$ are linearly independent, then $u_1,dots,u_n$ are linearly independent.




      Additionally, a small note on terminology: it is preferable to say that ${ u_1,dots,u_n }$ is a linearly independent subset of $U$, and that ${ T(1,0), T(0,1) }$ is not linearly independent over $mathbb{R}$.






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        Your example is a valid counterexample to the claim.
        So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),dots,T(u_n)$ are linearly independent, then $u_1,dots,u_n$ are linearly independent.




        Additionally, a small note on terminology: it is preferable to say that ${ u_1,dots,u_n }$ is a linearly independent subset of $U$, and that ${ T(1,0), T(0,1) }$ is not linearly independent over $mathbb{R}$.






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          Your example is a valid counterexample to the claim.
          So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),dots,T(u_n)$ are linearly independent, then $u_1,dots,u_n$ are linearly independent.




          Additionally, a small note on terminology: it is preferable to say that ${ u_1,dots,u_n }$ is a linearly independent subset of $U$, and that ${ T(1,0), T(0,1) }$ is not linearly independent over $mathbb{R}$.






          share|cite|improve this answer











          $endgroup$



          Your example is a valid counterexample to the claim.
          So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),dots,T(u_n)$ are linearly independent, then $u_1,dots,u_n$ are linearly independent.




          Additionally, a small note on terminology: it is preferable to say that ${ u_1,dots,u_n }$ is a linearly independent subset of $U$, and that ${ T(1,0), T(0,1) }$ is not linearly independent over $mathbb{R}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 18:34

























          answered Nov 28 '18 at 17:21









          BrahadeeshBrahadeesh

          6,24242361




          6,24242361























              3












              $begingroup$

              You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
              $$alpha_1 T(u_1) + cdots + alpha_n T(u_n) = 0$$
              linearity tells us
              $$T(alpha_1 u_1 + cdots + alpha_n u_n) = 0,$$
              and hence if we apply $T^{-1}$ to both sides
              $$alpha_1 u_1 + cdots + alpha_n u_n = 0.$$ Since ${u_1,ldots,u_n}$ are linearly independent this implies that $alpha_1=cdots=alpha_n=0$, and then by definition ${T(u_1),ldots,T(u_n)}$ are linearly independent as well.






              share|cite|improve this answer









              $endgroup$









              • 3




                $begingroup$
                Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
                $endgroup$
                – Thomas Andrews
                Nov 28 '18 at 17:27
















              3












              $begingroup$

              You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
              $$alpha_1 T(u_1) + cdots + alpha_n T(u_n) = 0$$
              linearity tells us
              $$T(alpha_1 u_1 + cdots + alpha_n u_n) = 0,$$
              and hence if we apply $T^{-1}$ to both sides
              $$alpha_1 u_1 + cdots + alpha_n u_n = 0.$$ Since ${u_1,ldots,u_n}$ are linearly independent this implies that $alpha_1=cdots=alpha_n=0$, and then by definition ${T(u_1),ldots,T(u_n)}$ are linearly independent as well.






              share|cite|improve this answer









              $endgroup$









              • 3




                $begingroup$
                Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
                $endgroup$
                – Thomas Andrews
                Nov 28 '18 at 17:27














              3












              3








              3





              $begingroup$

              You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
              $$alpha_1 T(u_1) + cdots + alpha_n T(u_n) = 0$$
              linearity tells us
              $$T(alpha_1 u_1 + cdots + alpha_n u_n) = 0,$$
              and hence if we apply $T^{-1}$ to both sides
              $$alpha_1 u_1 + cdots + alpha_n u_n = 0.$$ Since ${u_1,ldots,u_n}$ are linearly independent this implies that $alpha_1=cdots=alpha_n=0$, and then by definition ${T(u_1),ldots,T(u_n)}$ are linearly independent as well.






              share|cite|improve this answer









              $endgroup$



              You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
              $$alpha_1 T(u_1) + cdots + alpha_n T(u_n) = 0$$
              linearity tells us
              $$T(alpha_1 u_1 + cdots + alpha_n u_n) = 0,$$
              and hence if we apply $T^{-1}$ to both sides
              $$alpha_1 u_1 + cdots + alpha_n u_n = 0.$$ Since ${u_1,ldots,u_n}$ are linearly independent this implies that $alpha_1=cdots=alpha_n=0$, and then by definition ${T(u_1),ldots,T(u_n)}$ are linearly independent as well.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 28 '18 at 17:22









              cdipaolocdipaolo

              665312




              665312








              • 3




                $begingroup$
                Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
                $endgroup$
                – Thomas Andrews
                Nov 28 '18 at 17:27














              • 3




                $begingroup$
                Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
                $endgroup$
                – Thomas Andrews
                Nov 28 '18 at 17:27








              3




              3




              $begingroup$
              Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
              $endgroup$
              – Thomas Andrews
              Nov 28 '18 at 17:27




              $begingroup$
              Yes, being injective is the same as having a left-inverse. (That's easy to show for finite-dimensional $U,V$, but might require the axiom of choice in the case of infinite dimensions?)
              $endgroup$
              – Thomas Andrews
              Nov 28 '18 at 17:27











              3












              $begingroup$

              In general, this isn't true as your counterexample shows.



              If $T$ is invertible then the statement holds. Start from
              $$c_1T(u_1)+cdots+c_nT(u_n)=0$$
              Apply $T^{-1}$ on both sides and get
              $$c_1u_1+cdots+c_nu_n=0$$
              and $c_1=cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                In general, this isn't true as your counterexample shows.



                If $T$ is invertible then the statement holds. Start from
                $$c_1T(u_1)+cdots+c_nT(u_n)=0$$
                Apply $T^{-1}$ on both sides and get
                $$c_1u_1+cdots+c_nu_n=0$$
                and $c_1=cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  In general, this isn't true as your counterexample shows.



                  If $T$ is invertible then the statement holds. Start from
                  $$c_1T(u_1)+cdots+c_nT(u_n)=0$$
                  Apply $T^{-1}$ on both sides and get
                  $$c_1u_1+cdots+c_nu_n=0$$
                  and $c_1=cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.






                  share|cite|improve this answer









                  $endgroup$



                  In general, this isn't true as your counterexample shows.



                  If $T$ is invertible then the statement holds. Start from
                  $$c_1T(u_1)+cdots+c_nT(u_n)=0$$
                  Apply $T^{-1}$ on both sides and get
                  $$c_1u_1+cdots+c_nu_n=0$$
                  and $c_1=cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 17:23









                  Olivier MoschettaOlivier Moschetta

                  2,8311411




                  2,8311411






























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