Linear independence and basis within linear transformations












0












$begingroup$


I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:



Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.




  1. If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent

  2. If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.


My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...



Am I on the right path?



Thank you!






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:



    Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.




    1. If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent

    2. If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.


    My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...



    Am I on the right path?



    Thank you!






    linear-algebra linear-transformations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:



      Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.




      1. If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent

      2. If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.


      My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...



      Am I on the right path?



      Thank you!






      linear-algebra linear-transformations







      share|cite|improve this question











      $endgroup$




      I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:



      Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.




      1. If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent

      2. If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.


      My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...



      Am I on the right path?



      Thank you!






      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 Jan 1 at 18:38









      Hans Hüttel

      3,3572921




      3,3572921










      asked Jan 1 at 18:32









      daltadalta

      1508




      1508






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$


          1. It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.

          2. It is false, but you should provide a counterexample.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Your answers are correct but your reasoning is invalid.



            As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.



            So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.



            The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.



            For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.






            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%2f3058733%2flinear-independence-and-basis-within-linear-transformations%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









              1












              $begingroup$


              1. It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.

              2. It is false, but you should provide a counterexample.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$


                1. It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.

                2. It is false, but you should provide a counterexample.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  1. It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.

                  2. It is false, but you should provide a counterexample.






                  share|cite|improve this answer









                  $endgroup$




                  1. It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.

                  2. It is false, but you should provide a counterexample.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 18:39









                  José Carlos SantosJosé Carlos Santos

                  171k23132240




                  171k23132240























                      2












                      $begingroup$

                      Your answers are correct but your reasoning is invalid.



                      As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.



                      So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.



                      The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.



                      For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        Your answers are correct but your reasoning is invalid.



                        As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.



                        So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.



                        The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.



                        For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Your answers are correct but your reasoning is invalid.



                          As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.



                          So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.



                          The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.



                          For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.






                          share|cite|improve this answer











                          $endgroup$



                          Your answers are correct but your reasoning is invalid.



                          As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.



                          So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.



                          The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.



                          For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 1 at 18:45

























                          answered Jan 1 at 18:39









                          Ben WBen W

                          2,321615




                          2,321615






























                              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%2f3058733%2flinear-independence-and-basis-within-linear-transformations%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

                              Probability when a professor distributes a quiz and homework assignment to a class of n students.

                              Image
                              0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                              Read more

                              Aardman Animations

                              Image
                              Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                              Read more

                              Are they similar matrix

                              Image
                              0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                              Read more