Let T : M 2 x 2 → R 3 have matrix [ T ] B , A = [ 1 2 0 1 0 1 1 0 1 1 − 1 − 1 ] relative to A = { [ 2 0 0 0 ] , [ 0 3 0 0 ] , [ 0 0 5 0 ] , [ 0 0 0 6 ] } and β = { ( 1 , 1 , 1 ) , ( 1 , 2 , 3 ) , ( 1 , 4 , 9 ) }. Find the matrix of T relative to the bases A ′ = { [ 1 0 0 0 ] , [ 0 4 0 0 ] , [ 0 0 2 0 ] , [ 0 0 0 7 ] } and β ′ = { ( 1 , 1 , 1 ) , ( 1 , 0 , 0 ) , ( 1 , 1 , 0 ) }

Question

Let   T  :      M          2      x      2        →        R    3   have matrix   [  T      ]          B      ,      A        =      [                            1                          2                          0                          1                                      0                          1                          1                          0                                      1                          1                          −          1                          −          1                      ]   relative to   A  =  {      [                            2                          0                                      0                          0                      ]    ,      [                            0                          3                                      0                          0                      ]    ,      [                            0                          0                                      5                          0                      ]    ,      [                            0                          0                                      0                          6                      ]    } and   β  =  {  (  1  ,  1  ,  1  )  ,  (  1  ,  2  ,  3  )  ,  (  1  ,  4  ,  9  )  }. Find the matrix of T relative to the bases       A    ′    =  {      [                            1                          0                                      0                          0                      ]    ,      [                            0                          4                                      0                          0                      ]    ,      [                            0                          0                                      2                          0                      ]    ,      [                            0                          0                                      0                          7                      ]    } and       β    ′    =  {  (  1  ,  1  ,  1  )  ,  (  1  ,  0  ,  0  )  ,  (  1  ,  1  ,  0  )  }

Kaya Kemp · Accepted Answer

First of all, notice that       M          2      x      2        ≅      R    4   with       [                            a                          b                                      c                          d                      ]    →  (  a  ,  b  ,  c  ,  d  ).Your first step is the right way to go, but I would be careful with the notation. Let   α be the standard basis auf         R    3  , then  T  (      [                            2                          0                                      0                          0                      ]    )  =  (  1  ,  0  ,  1      )    B    =  (  2  ,  5  ,  10      )    α    =  (  10  ,  −  3  ,  −  5      )            B      ′        T  (      [                            0                          3                                      0                          0                      ]    )  =  (  2  ,  1  ,  1      )    B    =  (  4  ,  8  ,  9      )    α    =  (  9  ,  −  4  ,  −  1      )            B      ′        T  (      [                            0                          0                                      5                          0                      ]    )  =  (  0  ,  1  ,  −  1      )    B    =  (  2  ,  5  ,  12      )    α    =  (  12  ,  −  3  ,  −  7      )            B      ′        T  (      [                            0                          0                                      0                          6                      ]    )  =  (  1  ,  0  ,  −  1      )    B    =  (  0  ,  −  3  ,  −  8      )    α    =  (  −  8  ,  3  ,  5      )            B      ′      Thus, we have  [  T      ]                  B        ′            ,      A        =      [                            10                          9                          12                          −          8                                      −          3                          −          4                          −          3                          3                                      −          5                          −          1                          −          7                          5                      ]  Now we are almost done. The new basis       A    ′   is good to handle because every new basis vector is a scalar multiple of a vector of    A. For example the first basis vector of       A    ′   is half of its counterpart of   A (So   (  2  ,  0  ,  0  ,  0      )            A      ′        =  (  1  ,  0  ,  0  ,  0      )    A  ). So if you map   (  1  ,  0  ,  0  ,  0      )            A      ′       with our wanted matrix, it will have the same result as if we map   (      1    2    ,  0  ,  0  ,  0      )    A   into   [  T      ]                  B        ′            ,      A       (because they both map into the vector space generated by       B    ′  ) Thus, the first column of   [  T      ]                  B        ′            ,              A        ′            . You can get the other columns with the same method. The solution is:  [  T      ]                  B        ′            ,              A        ′              ==      [                            5                          9                      4            3                                    12                      2            5                                    −          8                      7            6                                                −                      3            2                                    −          4                      4            3                                    −          3                      2            5                                    3                      7            6                                                            5            2                                    −          1                      4            3                                    −          7                      2            5                                    5                      7            6                                ]

Let T : M 2 x 2 </mrow> </msub> →<!-- \rightarrow

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