T ( p ( x ) ) = ∫ 0 1 p ( x ) d x .(a) Show T is a linear transformation.(b) Compute N ( T ) . Is T one-to-one?(c) Show that T is onto.(d) Let B be the standard basis for P 2 and let B ′ = { 1 } be a basis for R. Find [ T ] B B ′ .(e) Use the matrix found in part (d) to compute T ( − x 2 − 3 x + 2 )

Name: T ( p ( x ) ) = ∫ 0 1 p ( x ) d x .(a) Show T is a linear transformation.(b) Compute N ( T ) . Is T one-to-one?(c) Show that T is onto.(d) Let B be the standard basis for P 2 and let B ′ = { 1 } be a basis for R. Find [ T ] B B ′ .(e) Use the matrix found in part (d) to compute T ( − x 2 − 3 x + 2 )
Uploaded: 2022-02-23T17:22:57+00:00
Description: T ( p ( x ) ) = ∫ 0 1 p ( x ) d x .(a) Show T is a linear transformation.(b) Compute N ( T ) . Is T one-to-one?(c) Show that T is onto.(d) Let B be the standard basis for P 2 and let B ′ = { 1 } be a basis for R. Find [ T ] B B ′ .(e) Use the matrix found in part (d) to compute T ( − x 2 − 3 x + 2 )

Question

T  (  p  (  x  )  )  =      ∫      0      1    p  (  x  )  d  x  .(a) Show   T is a linear transformation.(b) Compute   N  (  T  )  . Is   T one-to-one?(c) Show that   T is onto.(d) Let   B be the standard basis for         P      2   and let       B      ′    =  {  1  } be a basis for   R. Find   [  T      ]      B            B          ′      .(e) Use the matrix found in part (d) to compute   T      (    −          x          2        −    3    x    +    2    )

Nicolas Calhoun · Accepted Answer

(b) As explained in the comment above,  ker  ⁡  T  =      {          a          0        +          a          1        x    +          a          2              x      2        ∈              R          2        [    x    ]    :          a          0        +                  a              1            2        +                  a              2            3        =    0    }  We see, for instance, that   1  +  x  −      9    2        x      2    ∈  ker  ⁡  T, so   T is not injective.(c) To see that   T is onto, let   r  ∈  R be arbitrary and consider the constant polynomial   p  (  x  )  =  r. Then   T  p  =  r.(d) To find   [  T      ]      B            B      ′      , we take the basis   1  ,  x  ,      x      2   and evaluate   T at each of these polynomials. We then get   1  ,      1    2    ,      1    3  , so  [  T      ]      B            B      ′        =      (    1              1      2                  1      3        )    .(e) Finally,   T  (  −      x      2    −  3  x  +  2  )  =  −  T  (      x      2    )  −  3  T  (  x  )  +  2  T  (  1  )  =  −      1    3    −      3    2    +  2  =      1    6    .