Prove that there exists a non-zero vector $v in V$ such that $v perp U$












2












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$DeclareMathOperator{dim}{dim}$In Euclidean space $(X,leftlangle cdot , cdot rightrangle )$ where $dim X < infty$. Suppose we have two subspaces $U,V subset X$ and $dim U<dim V$. Prove that there exists a non-zero vector $v in V$ such that $v perp U$.




I tried to do this task however I don't have any idea how to prove it because I have less information about $U$ and I cannot even use the condition for the perpendicularity of vectors because it does not give me anything.



Can you help me?






linear-algebra







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    $begingroup$



    $DeclareMathOperator{dim}{dim}$In Euclidean space $(X,leftlangle cdot , cdot rightrangle )$ where $dim X < infty$. Suppose we have two subspaces $U,V subset X$ and $dim U<dim V$. Prove that there exists a non-zero vector $v in V$ such that $v perp U$.




    I tried to do this task however I don't have any idea how to prove it because I have less information about $U$ and I cannot even use the condition for the perpendicularity of vectors because it does not give me anything.



    Can you help me?






    linear-algebra







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      $DeclareMathOperator{dim}{dim}$In Euclidean space $(X,leftlangle cdot , cdot rightrangle )$ where $dim X < infty$. Suppose we have two subspaces $U,V subset X$ and $dim U<dim V$. Prove that there exists a non-zero vector $v in V$ such that $v perp U$.




      I tried to do this task however I don't have any idea how to prove it because I have less information about $U$ and I cannot even use the condition for the perpendicularity of vectors because it does not give me anything.



      Can you help me?






      linear-algebra







      share|cite|improve this question











      $endgroup$





      $DeclareMathOperator{dim}{dim}$In Euclidean space $(X,leftlangle cdot , cdot rightrangle )$ where $dim X < infty$. Suppose we have two subspaces $U,V subset X$ and $dim U<dim V$. Prove that there exists a non-zero vector $v in V$ such that $v perp U$.




      I tried to do this task however I don't have any idea how to prove it because I have less information about $U$ and I cannot even use the condition for the perpendicularity of vectors because it does not give me anything.



      Can you help me?






      linear-algebra




      linear-algebra






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 1:25









      Clayton

      19.6k33288




      19.6k33288










      asked Feb 2 at 0:53









      MP3129MP3129

      802211




      802211






















          3 Answers
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          $begingroup$

          Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$, the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{mid V}+dim Imp_{mid V}$ since $dim Imp_{mid V}<dim U$ this implies that $dim kerp_{mid V}>0$ since $dimU<dim V$), there exists $vin V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.






          share|cite|improve this answer









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            2












            $begingroup$

            We have that $U^perp$ is a subspace of $X$ whose codimension is $dim(U)$, while the codimension of $V$ is $dim(X) - dim(V)$. Since the sum of these two codimensions is $dim(X) - dim(V) + dim(U) < dim(X)$, it follows that $U^perp cap V$ is not the trivial subspace. Therefore, if you take a nonzero member $v in U^perp cap V$, then $v in V$ and $v$ is orthogonal to $U$.






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              A somewhat more concrete way to see this is as follows:



              since



              $dim X < infty, tag 1$



              we have



              $dim U < infty tag 2$



              as well; let



              $vec e_i in U, ; 1 le i le dim U, tag 3$



              be an orthonormal basis (for $U$, of course). For $vec x in X$ define



              $Pvec x = displaystyle sum_1^{dim U} langle vec x, vec e_i rangle vec e_i; tag 4$



              it is easy to see that



              $P:X to U tag 5$



              is a linear map; also, for



              $vec y in U, tag 6$



              we have



              $vec y = displaystyle sum_1^{dim U} langle vec y, vec e_i rangle vec e_i, tag 7$



              thus, since $langle vec e_i, vec e_j rangle = delta_{ij}$,



              $Pvec y = displaystyle sum_{i = 1}^{dim U} left langle sum_{j = 1}^{dim U} langle vec y, vec e_j rangle vec e_j, vec e_i right rangle vec e_i = sum_1^{dim U} langle vec y, vec e_i rangle vec e_i = vec y; tag 8$



              that is, $P$ fixes $U$, element-wise. Thus, in light of (5),



              $vec x in X Longrightarrow P^2 vec x = P(P vec x) = P vec x, tag 9$



              i.e.,



              $P^2 = P, tag{10}$



              a projection operator on $X$, with range $U$; a projection onto $U$. Furthermore, it is also easy to see that $P$ is self-adjoint, that is,



              $vec x, vec y in X Longrightarrow langle P vec x, vec y rangle = langle vec x, P vec y rangle; tag{11}$



              indeed,



              $langle P vec x, vec y rangle = left langle displaystyle sum_1^{dim U} langle vec x, vec e_j rangle vec e_j, vec y right rangle = displaystyle sum_1^{dim U} langle vec x, vec e_j rangle langle vec e_j, vec y rangle = sum_1^{dim U} langle vec e_j, vec y rangle langle vec x, vec e_j rangle$
              $= displaystyle sum_1^{dim U} langle vec y, vec e_j rangle langle vec e_j, vec x rangle = left langle displaystyle sum_1^{dim U} langle vec y, vec e_j rangle vec e_j, vec x right rangle = langle Pvec y, vec x rangle = langle vec x, Pvec y rangle; tag{12}$



              now with (10) and (11) in hand we may resolve the primary question as follows: since $dim U < dim V$ we may choose



              $vec v in V setminus U; tag{13}$



              then since



              $vec v notin U, tag{14}$



              it follows that



              $vec v ne P vec v in U; tag{15}$



              thus



              $vec v - Pvec v ne 0; tag{16}$



              also,



              $vec u in U Longrightarrow langle vec u, vec v - Pvec v rangle = langle vec u, vec v rangle - langle vec u, P vec v rangle = langle vec u, vec v rangle - langle P vec u,vec v rangle = langle vec u, vec v rangle - langle vec u, vec v rangle = 0, tag{17}$



              and we conclude that



              $0 ne vec v - P vec v in U^bot tag{18}$



              as required.






              share|cite|improve this answer











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                3 Answers
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                3 Answers
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                active

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                active

                oldest

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                active

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                2












                $begingroup$

                Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$, the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{mid V}+dim Imp_{mid V}$ since $dim Imp_{mid V}<dim U$ this implies that $dim kerp_{mid V}>0$ since $dimU<dim V$), there exists $vin V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$, the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{mid V}+dim Imp_{mid V}$ since $dim Imp_{mid V}<dim U$ this implies that $dim kerp_{mid V}>0$ since $dimU<dim V$), there exists $vin V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$, the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{mid V}+dim Imp_{mid V}$ since $dim Imp_{mid V}<dim U$ this implies that $dim kerp_{mid V}>0$ since $dimU<dim V$), there exists $vin V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.






                    share|cite|improve this answer









                    $endgroup$



                    Let $p$ be the orthogonal projection on $U$, since $dimU <dim V$, the kernel of the restriction of $p$ to $V$ is not zero ($dim V=dim ker p_{mid V}+dim Imp_{mid V}$ since $dim Imp_{mid V}<dim U$ this implies that $dim kerp_{mid V}>0$ since $dimU<dim V$), there exists $vin V$ such that $p(v)=0$, this implies that $v$ is orthogonal to $U$.







                    share|cite|improve this answer












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                    share|cite|improve this answer










                    answered Feb 2 at 0:58









                    Tsemo AristideTsemo Aristide

                    60.5k11446




                    60.5k11446























                        2












                        $begingroup$

                        We have that $U^perp$ is a subspace of $X$ whose codimension is $dim(U)$, while the codimension of $V$ is $dim(X) - dim(V)$. Since the sum of these two codimensions is $dim(X) - dim(V) + dim(U) < dim(X)$, it follows that $U^perp cap V$ is not the trivial subspace. Therefore, if you take a nonzero member $v in U^perp cap V$, then $v in V$ and $v$ is orthogonal to $U$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          We have that $U^perp$ is a subspace of $X$ whose codimension is $dim(U)$, while the codimension of $V$ is $dim(X) - dim(V)$. Since the sum of these two codimensions is $dim(X) - dim(V) + dim(U) < dim(X)$, it follows that $U^perp cap V$ is not the trivial subspace. Therefore, if you take a nonzero member $v in U^perp cap V$, then $v in V$ and $v$ is orthogonal to $U$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            We have that $U^perp$ is a subspace of $X$ whose codimension is $dim(U)$, while the codimension of $V$ is $dim(X) - dim(V)$. Since the sum of these two codimensions is $dim(X) - dim(V) + dim(U) < dim(X)$, it follows that $U^perp cap V$ is not the trivial subspace. Therefore, if you take a nonzero member $v in U^perp cap V$, then $v in V$ and $v$ is orthogonal to $U$.






                            share|cite|improve this answer









                            $endgroup$



                            We have that $U^perp$ is a subspace of $X$ whose codimension is $dim(U)$, while the codimension of $V$ is $dim(X) - dim(V)$. Since the sum of these two codimensions is $dim(X) - dim(V) + dim(U) < dim(X)$, it follows that $U^perp cap V$ is not the trivial subspace. Therefore, if you take a nonzero member $v in U^perp cap V$, then $v in V$ and $v$ is orthogonal to $U$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 2 at 1:13









                            Daniel ScheplerDaniel Schepler

                            9,3341821




                            9,3341821























                                2












                                $begingroup$

                                A somewhat more concrete way to see this is as follows:



                                since



                                $dim X < infty, tag 1$



                                we have



                                $dim U < infty tag 2$



                                as well; let



                                $vec e_i in U, ; 1 le i le dim U, tag 3$



                                be an orthonormal basis (for $U$, of course). For $vec x in X$ define



                                $Pvec x = displaystyle sum_1^{dim U} langle vec x, vec e_i rangle vec e_i; tag 4$



                                it is easy to see that



                                $P:X to U tag 5$



                                is a linear map; also, for



                                $vec y in U, tag 6$



                                we have



                                $vec y = displaystyle sum_1^{dim U} langle vec y, vec e_i rangle vec e_i, tag 7$



                                thus, since $langle vec e_i, vec e_j rangle = delta_{ij}$,



                                $Pvec y = displaystyle sum_{i = 1}^{dim U} left langle sum_{j = 1}^{dim U} langle vec y, vec e_j rangle vec e_j, vec e_i right rangle vec e_i = sum_1^{dim U} langle vec y, vec e_i rangle vec e_i = vec y; tag 8$



                                that is, $P$ fixes $U$, element-wise. Thus, in light of (5),



                                $vec x in X Longrightarrow P^2 vec x = P(P vec x) = P vec x, tag 9$



                                i.e.,



                                $P^2 = P, tag{10}$



                                a projection operator on $X$, with range $U$; a projection onto $U$. Furthermore, it is also easy to see that $P$ is self-adjoint, that is,



                                $vec x, vec y in X Longrightarrow langle P vec x, vec y rangle = langle vec x, P vec y rangle; tag{11}$



                                indeed,



                                $langle P vec x, vec y rangle = left langle displaystyle sum_1^{dim U} langle vec x, vec e_j rangle vec e_j, vec y right rangle = displaystyle sum_1^{dim U} langle vec x, vec e_j rangle langle vec e_j, vec y rangle = sum_1^{dim U} langle vec e_j, vec y rangle langle vec x, vec e_j rangle$
                                $= displaystyle sum_1^{dim U} langle vec y, vec e_j rangle langle vec e_j, vec x rangle = left langle displaystyle sum_1^{dim U} langle vec y, vec e_j rangle vec e_j, vec x right rangle = langle Pvec y, vec x rangle = langle vec x, Pvec y rangle; tag{12}$



                                now with (10) and (11) in hand we may resolve the primary question as follows: since $dim U < dim V$ we may choose



                                $vec v in V setminus U; tag{13}$



                                then since



                                $vec v notin U, tag{14}$



                                it follows that



                                $vec v ne P vec v in U; tag{15}$



                                thus



                                $vec v - Pvec v ne 0; tag{16}$



                                also,



                                $vec u in U Longrightarrow langle vec u, vec v - Pvec v rangle = langle vec u, vec v rangle - langle vec u, P vec v rangle = langle vec u, vec v rangle - langle P vec u,vec v rangle = langle vec u, vec v rangle - langle vec u, vec v rangle = 0, tag{17}$



                                and we conclude that



                                $0 ne vec v - P vec v in U^bot tag{18}$



                                as required.






                                share|cite|improve this answer











                                $endgroup$


















                                  2












                                  $begingroup$

                                  A somewhat more concrete way to see this is as follows:



                                  since



                                  $dim X < infty, tag 1$



                                  we have



                                  $dim U < infty tag 2$



                                  as well; let



                                  $vec e_i in U, ; 1 le i le dim U, tag 3$



                                  be an orthonormal basis (for $U$, of course). For $vec x in X$ define



                                  $Pvec x = displaystyle sum_1^{dim U} langle vec x, vec e_i rangle vec e_i; tag 4$



                                  it is easy to see that



                                  $P:X to U tag 5$



                                  is a linear map; also, for



                                  $vec y in U, tag 6$



                                  we have



                                  $vec y = displaystyle sum_1^{dim U} langle vec y, vec e_i rangle vec e_i, tag 7$



                                  thus, since $langle vec e_i, vec e_j rangle = delta_{ij}$,



                                  $Pvec y = displaystyle sum_{i = 1}^{dim U} left langle sum_{j = 1}^{dim U} langle vec y, vec e_j rangle vec e_j, vec e_i right rangle vec e_i = sum_1^{dim U} langle vec y, vec e_i rangle vec e_i = vec y; tag 8$



                                  that is, $P$ fixes $U$, element-wise. Thus, in light of (5),



                                  $vec x in X Longrightarrow P^2 vec x = P(P vec x) = P vec x, tag 9$



                                  i.e.,



                                  $P^2 = P, tag{10}$



                                  a projection operator on $X$, with range $U$; a projection onto $U$. Furthermore, it is also easy to see that $P$ is self-adjoint, that is,



                                  $vec x, vec y in X Longrightarrow langle P vec x, vec y rangle = langle vec x, P vec y rangle; tag{11}$



                                  indeed,



                                  $langle P vec x, vec y rangle = left langle displaystyle sum_1^{dim U} langle vec x, vec e_j rangle vec e_j, vec y right rangle = displaystyle sum_1^{dim U} langle vec x, vec e_j rangle langle vec e_j, vec y rangle = sum_1^{dim U} langle vec e_j, vec y rangle langle vec x, vec e_j rangle$
                                  $= displaystyle sum_1^{dim U} langle vec y, vec e_j rangle langle vec e_j, vec x rangle = left langle displaystyle sum_1^{dim U} langle vec y, vec e_j rangle vec e_j, vec x right rangle = langle Pvec y, vec x rangle = langle vec x, Pvec y rangle; tag{12}$



                                  now with (10) and (11) in hand we may resolve the primary question as follows: since $dim U < dim V$ we may choose



                                  $vec v in V setminus U; tag{13}$



                                  then since



                                  $vec v notin U, tag{14}$



                                  it follows that



                                  $vec v ne P vec v in U; tag{15}$



                                  thus



                                  $vec v - Pvec v ne 0; tag{16}$



                                  also,



                                  $vec u in U Longrightarrow langle vec u, vec v - Pvec v rangle = langle vec u, vec v rangle - langle vec u, P vec v rangle = langle vec u, vec v rangle - langle P vec u,vec v rangle = langle vec u, vec v rangle - langle vec u, vec v rangle = 0, tag{17}$



                                  and we conclude that



                                  $0 ne vec v - P vec v in U^bot tag{18}$



                                  as required.






                                  share|cite|improve this answer











                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    A somewhat more concrete way to see this is as follows:



                                    since



                                    $dim X < infty, tag 1$



                                    we have



                                    $dim U < infty tag 2$



                                    as well; let



                                    $vec e_i in U, ; 1 le i le dim U, tag 3$



                                    be an orthonormal basis (for $U$, of course). For $vec x in X$ define



                                    $Pvec x = displaystyle sum_1^{dim U} langle vec x, vec e_i rangle vec e_i; tag 4$



                                    it is easy to see that



                                    $P:X to U tag 5$



                                    is a linear map; also, for



                                    $vec y in U, tag 6$



                                    we have



                                    $vec y = displaystyle sum_1^{dim U} langle vec y, vec e_i rangle vec e_i, tag 7$



                                    thus, since $langle vec e_i, vec e_j rangle = delta_{ij}$,



                                    $Pvec y = displaystyle sum_{i = 1}^{dim U} left langle sum_{j = 1}^{dim U} langle vec y, vec e_j rangle vec e_j, vec e_i right rangle vec e_i = sum_1^{dim U} langle vec y, vec e_i rangle vec e_i = vec y; tag 8$



                                    that is, $P$ fixes $U$, element-wise. Thus, in light of (5),



                                    $vec x in X Longrightarrow P^2 vec x = P(P vec x) = P vec x, tag 9$



                                    i.e.,



                                    $P^2 = P, tag{10}$



                                    a projection operator on $X$, with range $U$; a projection onto $U$. Furthermore, it is also easy to see that $P$ is self-adjoint, that is,



                                    $vec x, vec y in X Longrightarrow langle P vec x, vec y rangle = langle vec x, P vec y rangle; tag{11}$



                                    indeed,



                                    $langle P vec x, vec y rangle = left langle displaystyle sum_1^{dim U} langle vec x, vec e_j rangle vec e_j, vec y right rangle = displaystyle sum_1^{dim U} langle vec x, vec e_j rangle langle vec e_j, vec y rangle = sum_1^{dim U} langle vec e_j, vec y rangle langle vec x, vec e_j rangle$
                                    $= displaystyle sum_1^{dim U} langle vec y, vec e_j rangle langle vec e_j, vec x rangle = left langle displaystyle sum_1^{dim U} langle vec y, vec e_j rangle vec e_j, vec x right rangle = langle Pvec y, vec x rangle = langle vec x, Pvec y rangle; tag{12}$



                                    now with (10) and (11) in hand we may resolve the primary question as follows: since $dim U < dim V$ we may choose



                                    $vec v in V setminus U; tag{13}$



                                    then since



                                    $vec v notin U, tag{14}$



                                    it follows that



                                    $vec v ne P vec v in U; tag{15}$



                                    thus



                                    $vec v - Pvec v ne 0; tag{16}$



                                    also,



                                    $vec u in U Longrightarrow langle vec u, vec v - Pvec v rangle = langle vec u, vec v rangle - langle vec u, P vec v rangle = langle vec u, vec v rangle - langle P vec u,vec v rangle = langle vec u, vec v rangle - langle vec u, vec v rangle = 0, tag{17}$



                                    and we conclude that



                                    $0 ne vec v - P vec v in U^bot tag{18}$



                                    as required.






                                    share|cite|improve this answer











                                    $endgroup$



                                    A somewhat more concrete way to see this is as follows:



                                    since



                                    $dim X < infty, tag 1$



                                    we have



                                    $dim U < infty tag 2$



                                    as well; let



                                    $vec e_i in U, ; 1 le i le dim U, tag 3$



                                    be an orthonormal basis (for $U$, of course). For $vec x in X$ define



                                    $Pvec x = displaystyle sum_1^{dim U} langle vec x, vec e_i rangle vec e_i; tag 4$



                                    it is easy to see that



                                    $P:X to U tag 5$



                                    is a linear map; also, for



                                    $vec y in U, tag 6$



                                    we have



                                    $vec y = displaystyle sum_1^{dim U} langle vec y, vec e_i rangle vec e_i, tag 7$



                                    thus, since $langle vec e_i, vec e_j rangle = delta_{ij}$,



                                    $Pvec y = displaystyle sum_{i = 1}^{dim U} left langle sum_{j = 1}^{dim U} langle vec y, vec e_j rangle vec e_j, vec e_i right rangle vec e_i = sum_1^{dim U} langle vec y, vec e_i rangle vec e_i = vec y; tag 8$



                                    that is, $P$ fixes $U$, element-wise. Thus, in light of (5),



                                    $vec x in X Longrightarrow P^2 vec x = P(P vec x) = P vec x, tag 9$



                                    i.e.,



                                    $P^2 = P, tag{10}$



                                    a projection operator on $X$, with range $U$; a projection onto $U$. Furthermore, it is also easy to see that $P$ is self-adjoint, that is,



                                    $vec x, vec y in X Longrightarrow langle P vec x, vec y rangle = langle vec x, P vec y rangle; tag{11}$



                                    indeed,



                                    $langle P vec x, vec y rangle = left langle displaystyle sum_1^{dim U} langle vec x, vec e_j rangle vec e_j, vec y right rangle = displaystyle sum_1^{dim U} langle vec x, vec e_j rangle langle vec e_j, vec y rangle = sum_1^{dim U} langle vec e_j, vec y rangle langle vec x, vec e_j rangle$
                                    $= displaystyle sum_1^{dim U} langle vec y, vec e_j rangle langle vec e_j, vec x rangle = left langle displaystyle sum_1^{dim U} langle vec y, vec e_j rangle vec e_j, vec x right rangle = langle Pvec y, vec x rangle = langle vec x, Pvec y rangle; tag{12}$



                                    now with (10) and (11) in hand we may resolve the primary question as follows: since $dim U < dim V$ we may choose



                                    $vec v in V setminus U; tag{13}$



                                    then since



                                    $vec v notin U, tag{14}$



                                    it follows that



                                    $vec v ne P vec v in U; tag{15}$



                                    thus



                                    $vec v - Pvec v ne 0; tag{16}$



                                    also,



                                    $vec u in U Longrightarrow langle vec u, vec v - Pvec v rangle = langle vec u, vec v rangle - langle vec u, P vec v rangle = langle vec u, vec v rangle - langle P vec u,vec v rangle = langle vec u, vec v rangle - langle vec u, vec v rangle = 0, tag{17}$



                                    and we conclude that



                                    $0 ne vec v - P vec v in U^bot tag{18}$



                                    as required.







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                                    edited Feb 2 at 3:39

























                                    answered Feb 2 at 3:15









                                    Robert LewisRobert Lewis

                                    48.9k23168




                                    48.9k23168






























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