Find basis of solutions of this linear systemSupposed to find basis of the subspace of vector space R 3 of solutions of this linear system of equations: y = { x 1 + 2 x 2 − x 3 = 0 2 x 1 + 7 x 2 − 2 x 3 = 0 − x 1 + 3 x 2 + x 3 = 0 I solve this system and I got: x 1 = x 3 and x 2 = 0 x → = [ x 1 0 x 1 ] = x 1 [ 1 0 1 ] + 0 [ 0 0 0 ] Is the basis: [ 1 0 1 ] ?

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Find basis of solutions of this linear systemSupposed to find basis of the subspace of vector space         R      3   of solutions of this linear system of equations:  y  =      {                                        x                      1                    +          2                      x                      2                    −                      x                      3                    =          0                                      2                      x                      1                    +          7                      x                      2                    −          2                      x                      3                    =          0                                      −                      x                      1                    +          3                      x                      2                    +                      x                      3                    =          0                        I solve this system and I got:       x      1    =      x      3   and       x      2    =  0      x    →    =      [                                        x            1                                                0                                                  x            1                                ]    =      x      1        [                            1                                      0                                      1                      ]    +  0      [                            0                                      0                                      0                      ]  Is the basis:       [                            1                                      0                                      1                      ]  ?

thatuglygirlyu · Accepted Answer

The question was basically answered in comments, I will add an answer to remove it from the list of unanswered question.By adding the first and the third equation we get   5      x    2    =  0 which implies       x    2    =  0. If we now plug       x    2    =  0 into remaining equations, we see, that they are all multiples of       x    1    −      x    2    =  0.Hence all solutions are of the form   (      x    1    ,  0  ,      x    1    ). I.e., the solutions form the subspace   {  (      x    1    ,  0  ,      x    1    )  ;      x    1    ∈  R  }.Every vector in this subspace is a multiple of   (  1  ,  0  ,  1  ), which means that vector   (  1  ,  0  ,  1  ) generates the subspace.Hence this subspace is one-dimensional and basis consists of a single vector   (  1  ,  0  ,  1  ).Of course, we could take any non-zero multiple of   (  1  ,  0  ,  1  ) instead. For example, vector   (  −  2  ,  0  ,  −  2  ) generates the same subspace.

Find basis of solutions of this linear system Supposed to find basis of the subspace of vector spac

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