Let U ⊂ Q 4 be the subspace induced by the following vectors: u 1 = ( 2 2 6 0 ) , u 2 = ( 2 − 1 1 1 ) , u 3 = ( − 1 0 2 − 1 ) Compute the dimension of U and find a homogeneous system of linear equations over Q with as few equations as possible, such that its solution set is U.I've used Gaussian elimination and found out that U has dimension 3, hence u 1 , u 2 , u 3 form a basis of U.My problem: I haven't been able to find the according linear system and I have no idea how I should proceed with this.

Question

Let   U  ⊂            Q        4   be the subspace induced by the following vectors:      u    1    =      (                            2                                      2                                      6                                      0                      )    ,        u    2    =      (                            2                                      −          1                                      1                                      1                      )    ,        u    3    =      (                            −          1                                      0                                      2                                      −          1                      )  Compute the dimension of   U and find a homogeneous system of linear equations over       Q   with as few equations as possible, such that its solution set is   U.I&#039;ve used Gaussian elimination and found out that   U has dimension 3, hence       u    1    ,      u    2    ,      u    3   form a basis of   U.My problem: I haven&#039;t been able to find the according linear system and I have no idea how I should proceed with this.

Carmelo Payne · Accepted Answer

Since   dim  ⁡  U  =  3 and   dim  ⁡            Q        4    =  4,   U is a hyperplane, and therefore you can solve your problem with a system which consists of a single equation. One such equation is   −  3  x  +  z  +  5  t  =  0. Actually, any equation of the type   a  x  +  b  y  +  c  z  +  d  t  =  0 will do as long as      {                            2          a          +          2          b          +          6          c          =          0                                      2          a          −          b          +          c          +          d          =          0                                      −          a          +          2          c          −          d          =          0                        and   (  a  ,  b  ,  c  ,  d  )  ≠  (  0  ,  0  ,  0  ,  0  ) (that&#039;s how I got my solution).

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