Let B = { b 1 , b 2 } be the basis for R 2 with b 1 = [ 1 1 ] , b 2 = [ 0 1 ] . Furthermore, let T : R 2 → R 2 be a linear transformation. The matrix representation of T with respect to B is [ T ] B = [ 1 − 1 1 2 ] . What is the standard matrix of T?

Question

Let   B  =  {        b    1    ,        b    2    } be the basis for         R    2   with         b    1    =      [                            1                                      1                      ]  ,         b    2    =      [                            0                                      1                      ]  . Furthermore, let   T  :        R    2    →        R    2   be a linear transformation. The matrix representation of   T with respect to   B is   [  T      ]      B    =      [                            1                          −          1                                      1                          2                      ]  . What is the standard matrix of   T?

Cynthia Logan · Accepted Answer

You have      T    =          [                                                  t                          11                                                          t                          12                                                                          t                          21                                                          t                          22                                          ]        ,          B    =          [                                    1                                0                                                1                                1                              ]      The change of basis is  [  T      ]      B    =        B          −      1        T  B  =      [                            1                          −          1                                      1                          2                      ]  First step is to calculate         B          −      1      . A 2x2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,        M    =          [                                                  m                          11                                                          m                          12                                                                          m                          21                                                          m                          22                                          ]            ⟺                    M              −        1              =          1                        m                  11                          m                  22                −                  m                  12                          m                  21                            [                                                  m                          22                                            −                          m                          12                                                            −                          m                          21                                                          m                          11                                          ]      Applying this to   B, we get        B          −      1        =      [                            1                          0                                      −          1                          1                      ]  Thus, you need to solve      [                            1                          0                                      −          1                          1                      ]        [                                        t                      11                                                t                      12                                                            t                      21                                                t                      22                                ]        [                            1                          0                                      1                          1                      ]    =      [                            1                          −          1                                      1                          2                      ]  for       t      11  ,       t      12  ,       t      21  , and       t      22  .

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