Finding matrices which P is null space

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Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$



a) Find one base of subspace $P^bot$



b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.



a) I found base $(1,1,1,1)$ that is easy
b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are




A=$beginbmatrix
1& 1& 1& 1\
1& 1& 1& 1\
1 & 1& 1& 1\
1 & 1& 1& 1
endbmatrix$.








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    Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$



    a) Find one base of subspace $P^bot$



    b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.



    a) I found base $(1,1,1,1)$ that is easy
    b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are




    A=$beginbmatrix
    1& 1& 1& 1\
    1& 1& 1& 1\
    1 & 1& 1& 1\
    1 & 1& 1& 1
    endbmatrix$.








    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$



      a) Find one base of subspace $P^bot$



      b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.



      a) I found base $(1,1,1,1)$ that is easy
      b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are




      A=$beginbmatrix
      1& 1& 1& 1\
      1& 1& 1& 1\
      1 & 1& 1& 1\
      1 & 1& 1& 1
      endbmatrix$.








      share|cite|improve this question














      Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$



      a) Find one base of subspace $P^bot$



      b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.



      a) I found base $(1,1,1,1)$ that is easy
      b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are




      A=$beginbmatrix
      1& 1& 1& 1\
      1& 1& 1& 1\
      1 & 1& 1& 1\
      1 & 1& 1& 1
      endbmatrix$.










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      edited Aug 20 at 10:17









      Bernard

      111k635103




      111k635103










      asked Aug 20 at 10:15









      Marko Škorić

      876




      876




















          3 Answers
          3






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          Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.






          share|cite|improve this answer





























            up vote
            0
            down vote













            a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.



            b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.






            share|cite|improve this answer





























              up vote
              0
              down vote













              Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices



              $$beginbmatrix
              a& a& a& a\
              a& a& a& a\
              vdots& vdots& vdots& vdots\
              a & a& a& a
              endbmatrix$$



              with $a neq 0$.






              share|cite|improve this answer




















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                0
                down vote



                accepted










                Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.






                share|cite|improve this answer


























                  up vote
                  0
                  down vote



                  accepted










                  Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.






                  share|cite|improve this answer
























                    up vote
                    0
                    down vote



                    accepted







                    up vote
                    0
                    down vote



                    accepted






                    Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.






                    share|cite|improve this answer














                    Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Aug 20 at 11:39









                    Martin Sleziak

                    43.5k6113260




                    43.5k6113260










                    answered Aug 20 at 10:23









                    Kavi Rama Murthy

                    23.3k2933




                    23.3k2933




















                        up vote
                        0
                        down vote













                        a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.



                        b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.



                          b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.



                            b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.






                            share|cite|improve this answer














                            a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.



                            b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Aug 20 at 10:23

























                            answered Aug 20 at 10:17









                            5xum

                            82.4k383147




                            82.4k383147




















                                up vote
                                0
                                down vote













                                Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices



                                $$beginbmatrix
                                a& a& a& a\
                                a& a& a& a\
                                vdots& vdots& vdots& vdots\
                                a & a& a& a
                                endbmatrix$$



                                with $a neq 0$.






                                share|cite|improve this answer
























                                  up vote
                                  0
                                  down vote













                                  Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices



                                  $$beginbmatrix
                                  a& a& a& a\
                                  a& a& a& a\
                                  vdots& vdots& vdots& vdots\
                                  a & a& a& a
                                  endbmatrix$$



                                  with $a neq 0$.






                                  share|cite|improve this answer






















                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices



                                    $$beginbmatrix
                                    a& a& a& a\
                                    a& a& a& a\
                                    vdots& vdots& vdots& vdots\
                                    a & a& a& a
                                    endbmatrix$$



                                    with $a neq 0$.






                                    share|cite|improve this answer












                                    Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices



                                    $$beginbmatrix
                                    a& a& a& a\
                                    a& a& a& a\
                                    vdots& vdots& vdots& vdots\
                                    a & a& a& a
                                    endbmatrix$$



                                    with $a neq 0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Aug 20 at 10:24









                                    gimusi

                                    68.7k73685




                                    68.7k73685






















                                         

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