Show linear transformation is not one-to-one T ( x , y , z ) = [ x − 2 z − 3 x + y + 3 z x − y + z ] I've taken the matrix of coefficients: A = [ 1 0 − 2 − 3 1 3 1 − 1 1 ] And reduced it to row-echelon form as follows: A = [ 1 0 2 0 1 3 0 0 0 ] But, I'm not sure where to go from here in proving that this transformation is not one-to-one.I understand that a one-to-one transformation has a pivot in every column, but I'm not sure what this means and how it applies to this matrix.Where can I go from here in proving this?

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Show linear transformation is not one-to-one  T  (  x  ,  y  ,  z  )  =      [                            x          −          2          z                                      −          3          x          +          y          +          3          z                                      x          −          y          +          z                      ]  I&#039;ve taken the matrix of coefficients:  A  =      [                            1                          0                          −          2                                      −          3                          1                          3                                      1                          −          1                          1                      ]  And reduced it to row-echelon form as follows:  A  =      [                            1                          0                          2                                      0                          1                          3                                      0                          0                          0                      ]  But, I&#039;m not sure where to go from here in proving that this transformation is not one-to-one.I understand that a one-to-one transformation has a pivot in every column, but I&#039;m not sure what this means and how it applies to this matrix.Where can I go from here in proving this?

iletsa2ym · Accepted Answer

You should, instead, obtain the reduced A as:      A    ′    =      [                            1                          0                          −          2                                      0                          1                          −          3                                      0                          0                          0                      ]  From the reduced A, we notice that a general solution to   A            x      →        =            0      →        is given by:            x      →        =      [                            2          t                                      3          t                                      t                      ]  We also see that   A      [                            x                                      y                                      z                      ]    =  T  (  x  ,  y  ,  z  )Hence we obtain   T  (  2  t  ,  3  t  ,  t  )  =            0      →       for any t.This shows that T is not injective.In fact, the null space is       ⟨          [                                    2                                                3                                                1                              ]        ⟩

Show linear transformation is not one-to-one T(x,y,z)=[[x-2z],[-3x+y+3z],[x-y+z]] I've taken the matrix of coefficients: A=[[1,0,-2],[-3,1,3],[1,-1,1]] And reduced it to row-echelon form as follows: A=[[1,0,2],[0,1,3],[0,0,0]]

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