The linear transformation A : R 2 → R 2 is given by the images of basis vectors: A ( ( 1 , 1 ) ) = ( 2 , 1 ) and A ( ( 1 , 0 ) ) = ( 0 , 3 ).1. Find a matrix of linear transformation A in the basis ( 1 , 1 ) , ( 1 , 0 ).2. Find A ( ( 3 , 2 ) )3. Find vector x = ( x 1 , x 2 ) such that the matrix ( − 6 − 6 3 4 ) is matrix of the linear transformation A in the basis ( 0 , 3 )ю

Question

The linear transformation   A  :        R    2    →        R    2   is given by the images of basis vectors:   A  (  (  1  ,  1  )  )  =  (  2  ,  1  ) and   A  (  (  1  ,  0  )  )  =  (  0  ,  3  ).1. Find a matrix of linear transformation   A in the basis   (  1  ,  1  )  ,  (  1  ,  0  ).2. Find   A  (  (  3  ,  2  )  )3. Find vector   x  =  (      x    1    ,      x    2    ) such that the matrix       (                            −          6                          −          6                                      3                          4                      )   is matrix of the linear transformation   A in the basis   (  0  ,  3  )ю

Xzavier Shelton · Accepted Answer

If   B  =  {  v  ,  w  }  is an ordered basis of         R    2  , then the matrix representation,   M, of   A  :        R    2    →        R    2   with respect to this basis is the   2  ×  2 matrix that has as its first column the coordinates of   A  v with respect to   B and as its second column the coordinates of   A  w with respect to   BIf you write a vector   x in terms of this basis   x  =      c    1    v  +      c    2    w  ,then, setting,   [  x      ]      B    =  [            c      1              c      2        ]                    (1)                    [        A              x                  ]                  B                =        M        [              x                  ]                  B                .            That is, for   x written in the standard basis, the coordinates of   A  x with respect to   B are given by the product of the matrix   M with the coordinate matrix of   x with respect to   B.For part 1:The matrix representation of   A is easily found, since you were told what   A        (    (  1  ,  1  )        )   and   A        (    (  1  ,  0  )        )   were.We need to write   (  2  ,  1  ) and   (  0  ,  3  ) in terms of the basis   B  =  {  (  1  ,  1  )  ,  (  1  ,  0  )  }.  (  2  ,  1  )  =  1  (  1  ,  1  )  +  1  (  1  ,  0  )and   (  0  ,  3  )  =  3  (  1  ,  1  )  −  3  (  1  ,  0  )The matrix   M is  M  =        [                          1        1            ⏟              [      A      (      1      ,      1      )              ]              B                                     3                  −          3                    ⏟              [      A      (      1      ,      0      )              ]              B                      ]    .For part 2:You need to write   (  3  ,  2  ) in terms of   B:  (  3  ,  2  )  =  2  (  1  ,  1  )  +  1  (  1  ,  0  )  .Using the matrix representation of   A,  [  A        (    (  3  ,  2  )        )        ]      B    =        [          1    1           3          −      3                ]          [        2    1          ]    =        [        5          −      1              ]    .This gives the coordinates of   A  (  (  3  ,  2  )  ) with respect to   B, so  A  (  (  3  ,  2  )  )  =  5  (  1  ,  1  )  +  (  −  1  )  (  1  ,  0  )  =  (  4  ,  5  )  .For part 3: Let       B    ′    =  {  (      x    1    ,      x    2    )  ,  (  0  ,  3  )  }We know the matrix  W  =        [                −      6        3                 −      6        4            ]  is the matrix representation of   A with respect to       B    ′  .The second column of   W is   [  A        (    (  0  ,  3  )        )        ]            B      ′      . So,                    (2)                    A        (        (        0        ,        3        )        )        =        −        6        (                  x          1                ,                  x          2                )        +        4        (        0        ,        3        )        =        (        −        6                  x          1                ,        −        6                  x          2                +        12        )        .            But you can compute   A        (    (  0  ,  3  )        )   using the matrix representation from part 1.We find   [  A        (    (  0  ,  3  )        )        ]      B   first. Towards this end, we write   (  0  ,  3  ) in terms of the basis   B first. Solve:   (  0  ,  3  )  =      c    1    (  1  ,  1  )  +      c    2    (  1  ,  0  )to obtain                              c          1                            =        3                                      c          2                            =        −        3.            Then:  [  A        (    (  0  ,  3  )        )        ]      B    =        [          1    1           3          −      3                ]          [        3          −      3              ]    =        [              −      6        12          ]    .So, the coordinates of   A  (  (  0  ,  3  )  ) with respect to   B are   (  −  6  ,  12  ). So,                    (3)                    A        (        (        0        ,        3        )        )        =        −        6        (        1        ,        1        )        +        12        (        1        ,        0        )        =        (        6        ,        −        6        )            Comparing equations (2) and (3) gives                    6                    =        −        6                  x          1                                    −        6                    =        −        6                  x          2                +        12            This gives       x    1    =  −  1 and       x    2    =  3.

Dania Mueller · Accepted Answer

Are u sure for the part 1? Shouldn&#039;t I express images (2,1) and (0,3) over basis vectors (1,1), (1,0)? I got this: (2,1)=(1,1)+(1,0) (0,3)=3(1,1)-3(1,0) so the matrix of A in basis (1,1), (1,0) is [1 3; 1 -3].

The linear transformation A : <mi mathvariant="double-struck">R 2 </msup>

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