Does a family of linearly independent injective maps have a vector with linearly independent images?












13












$begingroup$


Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.



Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?



This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.



In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.






linear-algebra linear-transformations







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$endgroup$








  • 1




    $begingroup$
    Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
    $endgroup$
    – Rahul
    Dec 18 '18 at 5:57


















13












$begingroup$


Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.



Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?



This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.



In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
    $endgroup$
    – Rahul
    Dec 18 '18 at 5:57
















13












13








13


6



$begingroup$


Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.



Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?



This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.



In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$




Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.



Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?



This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.



In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.






linear-algebra linear-transformations




linear-algebra linear-transformations






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edited Dec 19 '18 at 14:21









Brahadeesh

6,46942363




6,46942363










asked Dec 17 '18 at 5:18









user277182user277182

456212




456212








  • 1




    $begingroup$
    Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
    $endgroup$
    – Rahul
    Dec 18 '18 at 5:57
















  • 1




    $begingroup$
    Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
    $endgroup$
    – Rahul
    Dec 18 '18 at 5:57










1




1




$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57






$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57












2 Answers
2






active

oldest

votes


















10





+50







$begingroup$

This fails for $n=3.$ Consider $f_i$ with matrix representations



$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$



Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.



    When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10





      +50







      $begingroup$

      This fails for $n=3.$ Consider $f_i$ with matrix representations



      $$
      begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
      begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
      begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
      $$



      Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$






      share|cite|improve this answer









      $endgroup$


















        10





        +50







        $begingroup$

        This fails for $n=3.$ Consider $f_i$ with matrix representations



        $$
        begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
        begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
        begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
        $$



        Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$






        share|cite|improve this answer









        $endgroup$
















          10





          +50







          10





          +50



          10




          +50



          $begingroup$

          This fails for $n=3.$ Consider $f_i$ with matrix representations



          $$
          begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
          begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
          begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
          $$



          Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$






          share|cite|improve this answer









          $endgroup$



          This fails for $n=3.$ Consider $f_i$ with matrix representations



          $$
          begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
          begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
          begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
          $$



          Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 14:03









          DapDap

          17.5k841




          17.5k841























              0












              $begingroup$

              Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.



              When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.



                When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.



                  When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.






                  share|cite|improve this answer









                  $endgroup$



                  Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.



                  When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 0:42









                  Jacob MaibachJacob Maibach

                  1,4802917




                  1,4802917






























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