T/F: The Range of a Linear Transformation must be a subset of the domain.












3












$begingroup$


I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
    $endgroup$
    – DonAntonio
    Jan 29 at 18:06






  • 2




    $begingroup$
    The range of a function is a subset of the codomain.
    $endgroup$
    – stressed out
    Jan 29 at 18:06






  • 3




    $begingroup$
    Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
    $endgroup$
    – Hyperion
    Jan 29 at 18:06
















3












$begingroup$


I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
    $endgroup$
    – DonAntonio
    Jan 29 at 18:06






  • 2




    $begingroup$
    The range of a function is a subset of the codomain.
    $endgroup$
    – stressed out
    Jan 29 at 18:06






  • 3




    $begingroup$
    Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
    $endgroup$
    – Hyperion
    Jan 29 at 18:06














3












3








3


0



$begingroup$


I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.






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










asked Jan 29 at 18:03









TrebondTrebond

463




463








  • 2




    $begingroup$
    Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
    $endgroup$
    – DonAntonio
    Jan 29 at 18:06






  • 2




    $begingroup$
    The range of a function is a subset of the codomain.
    $endgroup$
    – stressed out
    Jan 29 at 18:06






  • 3




    $begingroup$
    Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
    $endgroup$
    – Hyperion
    Jan 29 at 18:06














  • 2




    $begingroup$
    Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
    $endgroup$
    – DonAntonio
    Jan 29 at 18:06






  • 2




    $begingroup$
    The range of a function is a subset of the codomain.
    $endgroup$
    – stressed out
    Jan 29 at 18:06






  • 3




    $begingroup$
    Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
    $endgroup$
    – Hyperion
    Jan 29 at 18:06








2




2




$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06




$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06




2




2




$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06




$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06




3




3




$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06




$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06










2 Answers
2






active

oldest

votes


















3












$begingroup$

Consider the following example of linear transformation,




$T:mathbb R to {mathbb R}^2$



$T(x)= (x,0)$




Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.






    share|cite|improve this answer









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092523%2ft-f-the-range-of-a-linear-transformation-must-be-a-subset-of-the-domain%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Consider the following example of linear transformation,




      $T:mathbb R to {mathbb R}^2$



      $T(x)= (x,0)$




      Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Consider the following example of linear transformation,




        $T:mathbb R to {mathbb R}^2$



        $T(x)= (x,0)$




        Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Consider the following example of linear transformation,




          $T:mathbb R to {mathbb R}^2$



          $T(x)= (x,0)$




          Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).






          share|cite|improve this answer









          $endgroup$



          Consider the following example of linear transformation,




          $T:mathbb R to {mathbb R}^2$



          $T(x)= (x,0)$




          Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 18:15









          Mayuresh LMayuresh L

          1,331323




          1,331323























              2












              $begingroup$

              Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.






                  share|cite|improve this answer









                  $endgroup$



                  Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 29 at 18:11









                  José Carlos SantosJosé Carlos Santos

                  171k23132240




                  171k23132240






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092523%2ft-f-the-range-of-a-linear-transformation-must-be-a-subset-of-the-domain%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more