Find an Example for a linear map $T: F^4 to F^4$ such that $ImT = KerT = sp{(1, 1, 1, 1), (1, 1, 1, 0)}$












1












$begingroup$


Find an Example for a linear map $T: F^4 to F^4$



such that $$ImT = KerT = sp{(1, 1, 1, 1), (1, 1, 1, 0)}$$



My Attempt:



First I completed the two vectors ${(1, 1, 1, 1), (1, 1, 1, 0)}$ to a base of $F^4$, so I picked: ${(1, 0, 0, 0), (0, 0, 1, 0)}$



So I'm looking for a map T such that $$ T((1, 1, 1, 1) = T((1, 1, 1, 0)) = 0 ; T((1,0, 0, 0)) = (1, 1, 1, 1), T(( 0 , 0 , 1, 0)) = (1, 1, 1, 0) $$



Now, Let $(x, y, z, w) in F^4$.



This is where I got stuck. what does the vector $(x, y, z, w)$ needs to be existing in order that I could find a linear map $T$ as needed?






linear-algebra linear-transformations







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    1












    $begingroup$


    Find an Example for a linear map $T: F^4 to F^4$



    such that $$ImT = KerT = sp{(1, 1, 1, 1), (1, 1, 1, 0)}$$



    My Attempt:



    First I completed the two vectors ${(1, 1, 1, 1), (1, 1, 1, 0)}$ to a base of $F^4$, so I picked: ${(1, 0, 0, 0), (0, 0, 1, 0)}$



    So I'm looking for a map T such that $$ T((1, 1, 1, 1) = T((1, 1, 1, 0)) = 0 ; T((1,0, 0, 0)) = (1, 1, 1, 1), T(( 0 , 0 , 1, 0)) = (1, 1, 1, 0) $$



    Now, Let $(x, y, z, w) in F^4$.



    This is where I got stuck. what does the vector $(x, y, z, w)$ needs to be existing in order that I could find a linear map $T$ as needed?






    linear-algebra linear-transformations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Find an Example for a linear map $T: F^4 to F^4$



      such that $$ImT = KerT = sp{(1, 1, 1, 1), (1, 1, 1, 0)}$$



      My Attempt:



      First I completed the two vectors ${(1, 1, 1, 1), (1, 1, 1, 0)}$ to a base of $F^4$, so I picked: ${(1, 0, 0, 0), (0, 0, 1, 0)}$



      So I'm looking for a map T such that $$ T((1, 1, 1, 1) = T((1, 1, 1, 0)) = 0 ; T((1,0, 0, 0)) = (1, 1, 1, 1), T(( 0 , 0 , 1, 0)) = (1, 1, 1, 0) $$



      Now, Let $(x, y, z, w) in F^4$.



      This is where I got stuck. what does the vector $(x, y, z, w)$ needs to be existing in order that I could find a linear map $T$ as needed?






      linear-algebra linear-transformations







      share|cite|improve this question











      $endgroup$




      Find an Example for a linear map $T: F^4 to F^4$



      such that $$ImT = KerT = sp{(1, 1, 1, 1), (1, 1, 1, 0)}$$



      My Attempt:



      First I completed the two vectors ${(1, 1, 1, 1), (1, 1, 1, 0)}$ to a base of $F^4$, so I picked: ${(1, 0, 0, 0), (0, 0, 1, 0)}$



      So I'm looking for a map T such that $$ T((1, 1, 1, 1) = T((1, 1, 1, 0)) = 0 ; T((1,0, 0, 0)) = (1, 1, 1, 1), T(( 0 , 0 , 1, 0)) = (1, 1, 1, 0) $$



      Now, Let $(x, y, z, w) in F^4$.



      This is where I got stuck. what does the vector $(x, y, z, w)$ needs to be existing in order that I could find a linear map $T$ as needed?






      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 5 at 10:57







      Jneven

















      asked Jan 5 at 10:43









      JnevenJneven

      906322




      906322






















          2 Answers
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          $begingroup$

          Hint



          In the basis you chosed (I haven’t verified that it is indeed a basis), the matrix of a linear map satisfying the required condition is



          $$T=begin{pmatrix}
          0&0&1&0\
          0&0&0&1\
          0&0&0&0\
          0&0&0&0
          end{pmatrix}$$



          Then use a change of basis.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            $T(1,0,0,0)$ is the first column vector of $[T]$, while $T(0,0,1,0)$ is the third column vector. Thus, your second and third conditions imply that the first column of $[T]$ is $(1,1,1,1)^T$ and its second column is $(1,1,1,0)^T$.



            For $T(1,1,1,1)=T(1,1,1,0)=0$, you require that all rows of $[T]$ should be orthogonal to $(1,1,1,1),(1,1,1,0)$. Therefore,



            $$[T]=begin{bmatrix}1&-2&1&0\1&-2&1&0\1&-2&1&0\1&-1&0&0end{bmatrix}$$






            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









              2












              $begingroup$

              Hint



              In the basis you chosed (I haven’t verified that it is indeed a basis), the matrix of a linear map satisfying the required condition is



              $$T=begin{pmatrix}
              0&0&1&0\
              0&0&0&1\
              0&0&0&0\
              0&0&0&0
              end{pmatrix}$$



              Then use a change of basis.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint



                In the basis you chosed (I haven’t verified that it is indeed a basis), the matrix of a linear map satisfying the required condition is



                $$T=begin{pmatrix}
                0&0&1&0\
                0&0&0&1\
                0&0&0&0\
                0&0&0&0
                end{pmatrix}$$



                Then use a change of basis.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint



                  In the basis you chosed (I haven’t verified that it is indeed a basis), the matrix of a linear map satisfying the required condition is



                  $$T=begin{pmatrix}
                  0&0&1&0\
                  0&0&0&1\
                  0&0&0&0\
                  0&0&0&0
                  end{pmatrix}$$



                  Then use a change of basis.






                  share|cite|improve this answer









                  $endgroup$



                  Hint



                  In the basis you chosed (I haven’t verified that it is indeed a basis), the matrix of a linear map satisfying the required condition is



                  $$T=begin{pmatrix}
                  0&0&1&0\
                  0&0&0&1\
                  0&0&0&0\
                  0&0&0&0
                  end{pmatrix}$$



                  Then use a change of basis.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 10:52









                  mathcounterexamples.netmathcounterexamples.net

                  27k22157




                  27k22157























                      1












                      $begingroup$

                      $T(1,0,0,0)$ is the first column vector of $[T]$, while $T(0,0,1,0)$ is the third column vector. Thus, your second and third conditions imply that the first column of $[T]$ is $(1,1,1,1)^T$ and its second column is $(1,1,1,0)^T$.



                      For $T(1,1,1,1)=T(1,1,1,0)=0$, you require that all rows of $[T]$ should be orthogonal to $(1,1,1,1),(1,1,1,0)$. Therefore,



                      $$[T]=begin{bmatrix}1&-2&1&0\1&-2&1&0\1&-2&1&0\1&-1&0&0end{bmatrix}$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        $T(1,0,0,0)$ is the first column vector of $[T]$, while $T(0,0,1,0)$ is the third column vector. Thus, your second and third conditions imply that the first column of $[T]$ is $(1,1,1,1)^T$ and its second column is $(1,1,1,0)^T$.



                        For $T(1,1,1,1)=T(1,1,1,0)=0$, you require that all rows of $[T]$ should be orthogonal to $(1,1,1,1),(1,1,1,0)$. Therefore,



                        $$[T]=begin{bmatrix}1&-2&1&0\1&-2&1&0\1&-2&1&0\1&-1&0&0end{bmatrix}$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $T(1,0,0,0)$ is the first column vector of $[T]$, while $T(0,0,1,0)$ is the third column vector. Thus, your second and third conditions imply that the first column of $[T]$ is $(1,1,1,1)^T$ and its second column is $(1,1,1,0)^T$.



                          For $T(1,1,1,1)=T(1,1,1,0)=0$, you require that all rows of $[T]$ should be orthogonal to $(1,1,1,1),(1,1,1,0)$. Therefore,



                          $$[T]=begin{bmatrix}1&-2&1&0\1&-2&1&0\1&-2&1&0\1&-1&0&0end{bmatrix}$$






                          share|cite|improve this answer









                          $endgroup$



                          $T(1,0,0,0)$ is the first column vector of $[T]$, while $T(0,0,1,0)$ is the third column vector. Thus, your second and third conditions imply that the first column of $[T]$ is $(1,1,1,1)^T$ and its second column is $(1,1,1,0)^T$.



                          For $T(1,1,1,1)=T(1,1,1,0)=0$, you require that all rows of $[T]$ should be orthogonal to $(1,1,1,1),(1,1,1,0)$. Therefore,



                          $$[T]=begin{bmatrix}1&-2&1&0\1&-2&1&0\1&-2&1&0\1&-1&0&0end{bmatrix}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 at 11:05









                          Shubham JohriShubham Johri

                          5,204718




                          5,204718






























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