T ( p ( x ) ) = <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup> p

Kristen Stokes

Kristen Stokes

Answered question

2022-07-01

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 )

Answer & Explanation

Nicolas Calhoun

Nicolas Calhoun

Beginner2022-07-02Added 15 answers

(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 .

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