if $f$ is linear transformation, is proof that $f(U)$ is linear subspace and find its dim












1












$begingroup$


Given is Linear transformation $f in L(mathbb R^3, mathbb R[t]_3)$ such that:
$ker f = span([1,1,1]^T) $

Given is also $U = span([3,1,-1]^T,[-1,1,3]^T)$

Proof that $f(U)$ is linear subspace in $mathbb R[t]_3$ and find its dim



My try



$$ f(U) = left{ f(u) | u in U right} $$
Take a random $u in U$. Then
$$ u = alpha[3,1,-1]^T + beta[-1,1,3]^T$$
We know that $f$ is linear transformation so:
$$f(u) = alpha f left([3,1,-1]^T right) + beta fleft([-1,1,3]^Tright)$$



I check if the scalar multiplies or adds vectors from the subspace derives from $f(U)$



$$ gamma f(u) = (alphagamma) f left([3,1,-1]^T right) + (betagamma) fleft([-1,1,3]^Tright)$$
and
$$f(u)+f(u') = (alpha + alpha') f left([3,1,-1]^T right) + (beta + beta') fleft([-1,1,3]^Tright) $$
ok. So this is linear subspace.
Now I am going to find dim.
But $[3,1,-1]^T,[-1,1,3]^T $ are linearly independent. So
$$ dim f(U) = 2 $$



What is the problem?



The problem is that the correct answer should be
$$ dim f(U) = 1 $$
I suspect that it comes to $ker f = span([1,1,1]^T) $ because $ker f in U $. But why does it matter? I do not think that this is any theorem on that situation






linear-algebra proof-verification linear-transformations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Given is Linear transformation $f in L(mathbb R^3, mathbb R[t]_3)$ such that:
    $ker f = span([1,1,1]^T) $

    Given is also $U = span([3,1,-1]^T,[-1,1,3]^T)$

    Proof that $f(U)$ is linear subspace in $mathbb R[t]_3$ and find its dim



    My try



    $$ f(U) = left{ f(u) | u in U right} $$
    Take a random $u in U$. Then
    $$ u = alpha[3,1,-1]^T + beta[-1,1,3]^T$$
    We know that $f$ is linear transformation so:
    $$f(u) = alpha f left([3,1,-1]^T right) + beta fleft([-1,1,3]^Tright)$$



    I check if the scalar multiplies or adds vectors from the subspace derives from $f(U)$



    $$ gamma f(u) = (alphagamma) f left([3,1,-1]^T right) + (betagamma) fleft([-1,1,3]^Tright)$$
    and
    $$f(u)+f(u') = (alpha + alpha') f left([3,1,-1]^T right) + (beta + beta') fleft([-1,1,3]^Tright) $$
    ok. So this is linear subspace.
    Now I am going to find dim.
    But $[3,1,-1]^T,[-1,1,3]^T $ are linearly independent. So
    $$ dim f(U) = 2 $$



    What is the problem?



    The problem is that the correct answer should be
    $$ dim f(U) = 1 $$
    I suspect that it comes to $ker f = span([1,1,1]^T) $ because $ker f in U $. But why does it matter? I do not think that this is any theorem on that situation






    linear-algebra proof-verification linear-transformations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Given is Linear transformation $f in L(mathbb R^3, mathbb R[t]_3)$ such that:
      $ker f = span([1,1,1]^T) $

      Given is also $U = span([3,1,-1]^T,[-1,1,3]^T)$

      Proof that $f(U)$ is linear subspace in $mathbb R[t]_3$ and find its dim



      My try



      $$ f(U) = left{ f(u) | u in U right} $$
      Take a random $u in U$. Then
      $$ u = alpha[3,1,-1]^T + beta[-1,1,3]^T$$
      We know that $f$ is linear transformation so:
      $$f(u) = alpha f left([3,1,-1]^T right) + beta fleft([-1,1,3]^Tright)$$



      I check if the scalar multiplies or adds vectors from the subspace derives from $f(U)$



      $$ gamma f(u) = (alphagamma) f left([3,1,-1]^T right) + (betagamma) fleft([-1,1,3]^Tright)$$
      and
      $$f(u)+f(u') = (alpha + alpha') f left([3,1,-1]^T right) + (beta + beta') fleft([-1,1,3]^Tright) $$
      ok. So this is linear subspace.
      Now I am going to find dim.
      But $[3,1,-1]^T,[-1,1,3]^T $ are linearly independent. So
      $$ dim f(U) = 2 $$



      What is the problem?



      The problem is that the correct answer should be
      $$ dim f(U) = 1 $$
      I suspect that it comes to $ker f = span([1,1,1]^T) $ because $ker f in U $. But why does it matter? I do not think that this is any theorem on that situation






      linear-algebra proof-verification linear-transformations







      share|cite|improve this question









      $endgroup$




      Given is Linear transformation $f in L(mathbb R^3, mathbb R[t]_3)$ such that:
      $ker f = span([1,1,1]^T) $

      Given is also $U = span([3,1,-1]^T,[-1,1,3]^T)$

      Proof that $f(U)$ is linear subspace in $mathbb R[t]_3$ and find its dim



      My try



      $$ f(U) = left{ f(u) | u in U right} $$
      Take a random $u in U$. Then
      $$ u = alpha[3,1,-1]^T + beta[-1,1,3]^T$$
      We know that $f$ is linear transformation so:
      $$f(u) = alpha f left([3,1,-1]^T right) + beta fleft([-1,1,3]^Tright)$$



      I check if the scalar multiplies or adds vectors from the subspace derives from $f(U)$



      $$ gamma f(u) = (alphagamma) f left([3,1,-1]^T right) + (betagamma) fleft([-1,1,3]^Tright)$$
      and
      $$f(u)+f(u') = (alpha + alpha') f left([3,1,-1]^T right) + (beta + beta') fleft([-1,1,3]^Tright) $$
      ok. So this is linear subspace.
      Now I am going to find dim.
      But $[3,1,-1]^T,[-1,1,3]^T $ are linearly independent. So
      $$ dim f(U) = 2 $$



      What is the problem?



      The problem is that the correct answer should be
      $$ dim f(U) = 1 $$
      I suspect that it comes to $ker f = span([1,1,1]^T) $ because $ker f in U $. But why does it matter? I do not think that this is any theorem on that situation






      linear-algebra proof-verification linear-transformations




      linear-algebra proof-verification linear-transformations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 5 at 11:18









      VirtualUserVirtualUser

      1,303317




      1,303317






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ might not be linearly independent.



          Notice that $begin{bmatrix}3\1\-1end{bmatrix}=2begin{bmatrix}1\1\1end{bmatrix}-begin{bmatrix}-1\1\3end{bmatrix}$



          $f(u) =alpha f(begin{bmatrix}3&1&-1end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=alpha f(2begin{bmatrix}1&1&1end{bmatrix}^T-begin{bmatrix}-1&1&3end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=(beta-alpha)f(begin{bmatrix}-1&1&3end{bmatrix}^T)$



          Since $begin{bmatrix}-1&1&3end{bmatrix}^Tnotinker(f),f(begin{bmatrix}-1&1&3end{bmatrix}^T)ne0$. Thus, $dim f(U)=1$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?



            Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?



            Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.






            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%2f3062621%2fif-f-is-linear-transformation-is-proof-that-fu-is-linear-subspace-and-fin%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












              $begingroup$

              $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ might not be linearly independent.



              Notice that $begin{bmatrix}3\1\-1end{bmatrix}=2begin{bmatrix}1\1\1end{bmatrix}-begin{bmatrix}-1\1\3end{bmatrix}$



              $f(u) =alpha f(begin{bmatrix}3&1&-1end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=alpha f(2begin{bmatrix}1&1&1end{bmatrix}^T-begin{bmatrix}-1&1&3end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=(beta-alpha)f(begin{bmatrix}-1&1&3end{bmatrix}^T)$



              Since $begin{bmatrix}-1&1&3end{bmatrix}^Tnotinker(f),f(begin{bmatrix}-1&1&3end{bmatrix}^T)ne0$. Thus, $dim f(U)=1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ might not be linearly independent.



                Notice that $begin{bmatrix}3\1\-1end{bmatrix}=2begin{bmatrix}1\1\1end{bmatrix}-begin{bmatrix}-1\1\3end{bmatrix}$



                $f(u) =alpha f(begin{bmatrix}3&1&-1end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=alpha f(2begin{bmatrix}1&1&1end{bmatrix}^T-begin{bmatrix}-1&1&3end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=(beta-alpha)f(begin{bmatrix}-1&1&3end{bmatrix}^T)$



                Since $begin{bmatrix}-1&1&3end{bmatrix}^Tnotinker(f),f(begin{bmatrix}-1&1&3end{bmatrix}^T)ne0$. Thus, $dim f(U)=1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ might not be linearly independent.



                  Notice that $begin{bmatrix}3\1\-1end{bmatrix}=2begin{bmatrix}1\1\1end{bmatrix}-begin{bmatrix}-1\1\3end{bmatrix}$



                  $f(u) =alpha f(begin{bmatrix}3&1&-1end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=alpha f(2begin{bmatrix}1&1&1end{bmatrix}^T-begin{bmatrix}-1&1&3end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=(beta-alpha)f(begin{bmatrix}-1&1&3end{bmatrix}^T)$



                  Since $begin{bmatrix}-1&1&3end{bmatrix}^Tnotinker(f),f(begin{bmatrix}-1&1&3end{bmatrix}^T)ne0$. Thus, $dim f(U)=1$.






                  share|cite|improve this answer









                  $endgroup$



                  $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ might not be linearly independent.



                  Notice that $begin{bmatrix}3\1\-1end{bmatrix}=2begin{bmatrix}1\1\1end{bmatrix}-begin{bmatrix}-1\1\3end{bmatrix}$



                  $f(u) =alpha f(begin{bmatrix}3&1&-1end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=alpha f(2begin{bmatrix}1&1&1end{bmatrix}^T-begin{bmatrix}-1&1&3end{bmatrix}^T)+beta f(begin{bmatrix}-1&1&3end{bmatrix}^T)\=(beta-alpha)f(begin{bmatrix}-1&1&3end{bmatrix}^T)$



                  Since $begin{bmatrix}-1&1&3end{bmatrix}^Tnotinker(f),f(begin{bmatrix}-1&1&3end{bmatrix}^T)ne0$. Thus, $dim f(U)=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 11:52









                  Shubham JohriShubham Johri

                  5,550818




                  5,550818























                      1












                      $begingroup$

                      While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?



                      Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?



                      Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?



                        Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?



                        Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?



                          Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?



                          Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.






                          share|cite|improve this answer









                          $endgroup$



                          While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?



                          Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?



                          Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 at 11:28









                          Theo BenditTheo Bendit

                          20.5k12354




                          20.5k12354






























                              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%2f3062621%2fif-f-is-linear-transformation-is-proof-that-fu-is-linear-subspace-and-fin%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

                              Wiesbaden

                              Image
                              Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                              Read more

                              Marschland

                              Image
                              Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                              Read more

                              Dieringhausen

                              Image
                              Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                              Read more