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













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      asked Jan 5 at 11:18









      VirtualUserVirtualUser

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          2 Answers
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          $begingroup$

          $f(begin{bmatrix}3&1&-1end{bmatrix}^T),f(begin{bmatrix}-1&1&3end{bmatrix}^T)$ 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









            $endgroup$













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              2 Answers
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              2 Answers
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              active

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              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

                  5,204718




                  5,204718























                      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

                          19.2k12353




                          19.2k12353






























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