T ( x , y ) = ( 2 x − y , 3 x − y )with respect to basis (1,0), and (0,1) Now M(T)= [ 2 3 x − 1 − 1 y ] gives ( 2 x + 3 y − x − y ) so this cannot be the transformation matrixThe correct one is [ 2 − 1 x 3 − 1 y ] Also, how do I find the matrix transformation of (x,y,z)=(x+y,y+z) for basis (1,0) and (0,1). So we are going from 3 dimensions to 2. So my matrix should have 2 rows and 2 columns right?

Question

T  (  x  ,  y  )  =  (  2  x  −  y  ,  3  x  −  y  )with respect to basis (1,0), and (0,1) Now M(T)=      [                            2                          3                          x                                      −          1                          −          1                          y                      ]  gives      (                            2          x          +          3          y                                      −          x          −          y                      )  so this cannot be the transformation matrixThe correct one is      [                            2                          −          1                          x                                      3                          −          1                          y                      ]  Also, how do I find the matrix transformation of (x,y,z)=(x+y,y+z) for basis (1,0) and (0,1). So we are going from 3 dimensions to 2. So my matrix should have 2 rows and 2 columns right?

Smarmabagd7 · Accepted Answer

Consider linear map   T  :        R    3    →        R    2   defined by   T  (  x  ,  y  ,  z  )  =  (  x  +  y  ,  y  +  z  ), we always use the two bases for the domain and the co-domain, to obtain a matrix representation. For example, use the standard ordered bases for         R    3   and         R    2  , respectively. That is,  β  =  {      e    1    ,      e    2    ,      e    3    }  =      {          (                                    1                                                0                                                0                              )        ,          (                                    0                                                1                                                0                              )        ,          (                                    0                                                0                                                1                              )        }          and      γ  =  {              e      ¯        1    ,              e      ¯        2    }  =      {          (                                    1                                                0                              )        ,          (                                    0                                                1                              )        }    ,and do the following works:                          T        (                  e          1                )                            =                  (                                                    1                                                                    0                                              )                =        1        ⋅                                      e            ¯                    1                +        0        ⋅                                      e            ¯                    2                ,                                  T        (                  e          2                )                            =                  (                                                    1                                                                    1                                              )                =        1        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                ,                                  T        (                  e          3                )                            =                  (                                                    0                                                                    1                                              )                =        0        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                .                                      T        (                  e          1                )                            =                  (                                                    1                                                                    0                                              )                =        1        ⋅                                      e            ¯                    1                +        0        ⋅                                      e            ¯                    2                ,                                  T        (                  e          2                )                            =                  (                                                    1                                                                    1                                              )                =        1        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                ,                                  T        (                  e          3                )                            =                  (                                                    0                                                                    1                                              )                =        0        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                .                                      T        (                  e          1                )                            =                  (                                                    1                                                                    0                                              )                =        1        ⋅                                      e            ¯                    1                +        0        ⋅                                      e            ¯                    2                ,                                  T        (                  e          2                )                            =                  (                                                    1                                                                    1                                              )                =        1        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                ,                                  T        (                  e          3                )                            =                  (                                                    0                                                                    1                                              )                =        0        ⋅                                      e            ¯                    1                +        1        ⋅                                      e            ¯                    2                .            Hence we get the matrix representation of   T with respect to   β and   γ as below:  [  T      ]    β    γ    =      (                            1                          1                          0                                      0                          1                          1                      )    .

Charity Daniels · Accepted Answer

And what about the first one? Did I get that right?

T ( x , y ) = ( 2 x − y , 3 x −

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