A linear transformation between polynomial spaces f : P 2 ( R ) → P 2 ( R ) is given by f ( p ( x ) ) = 3 ⋅ p ( 1 ) − x 2 ⋅ p ( 0 ) + ( x − 1 ) ⋅ p ′ ( 1 )Determine the transformation matrix with respect to the monomial basis ( 1 , x , x 2 )

Question

A linear transformation between polynomial spaces   f  :      P    2    (  R  )  →      P    2    (  R  ) is given by  f  (  p  (  x  )  )  =  3  ⋅  p  (  1  )  −      x    2    ⋅  p  (  0  )  +  (  x  −  1  )  ⋅      p    ′    (  1  )Determine the transformation matrix with respect to the monomial basis  (  1  ,  x  ,      x    2    )

Trey Ross · Accepted Answer

If you have a basis   β  =  {  1  ,  x  ,      x    2    }, and we need to determine the matrix of   f wrt   β, then let&#039;s first write out   f in the way that makes it the most obvious how it&#039;s transforming it&#039;s coordinates:                    f        (        a                  x          2                +        b        x        +        c        )        =        f        (        (        a        ,        b        ,        c        )        )                            =        3        (        a        +        b        +        c        )        −        c                  x          2                +        (        2        a        +        b        )        x        −        (        2        a        +        b        )                                          =        (        −        c        ,        2        a        +        b        ,        3        (        a        +        b        +        c        )        −        (        2        a        +        b        )        )                                          =        (        −        c        ,        2        a        +        b        ,        a        +        2        b        +        3        c        )            I&#039;m using   (  a  ,  b  ,  c  ) to denote   a      x    2    +  b  x  +  c to emphasize the fact that these are coordinates in       P    2    (  R  ). So, if our matrix for   f is  [  f      ]    β    =      (                                        a                      11                                                a                      12                                                a                      13                                                            a                      21                                                a                      22                                                a                      23                                                            a                      31                                                a                      32                                                a                      33                                )  then notice that the first coordinate in   [  f      ]    β    (  a  ,  b  ,  c      )    T   is given by the dot product of the first row with   (  a  ,  b  ,  c  ), so       a      11    a  +      a      12    b  +      a      13    c  =  −  c, implying       a      13    =  −  1, and       a      11    =      a      12    =  0. We can continue this process for each row, since the ith coordinate of   [  f      ]    β    (  a  ,  b  ,  c      )    T   is given by the dot product of the   ith row with   (  a  ,  b  ,  c  ). Hence      a      21    a  +      a      22    b  +      a      23    c  =  2  a  +  b        a      31    a  +      a      32    b  +      a      33    c  =  a  +  2  b  +  3  cfrom which we can see that   (      a      21    ,      a      22    ,      a      23    )  =  (  2  ,  1  ,  0  ), and   (      a      31    ,      a      32    ,      a      33    )  =  (  1  ,  2  ,  3  ). Hence  [  f      ]    β    =      (                            0                          0                          −          1                                      2                          1                          0                                      1                          2                          3                      )  So you were on the right track, you just had your rows in the reverse order.

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