- #1

Benny

- 584

- 0

Q. Let P_2 be the vector space of all polynomials of degree at most 2. A function [tex]T:P_2 \to R^3 [/tex] is defined by [tex]T\left( {p\left( x \right)} \right) = \left( {p(1),p'(1),p''(1)} \right)[/tex]. That is, T evaluates the polynomial p(x) and its first two derivatives at the point x = 1.

a) Show that T is a linear transformation.

b) Find the matrix of T using the basis B = {1,x,x^2} for P_2 and the standard basis S = {(1,0,0),(0,1,0),(0,0,1)} for R^3.

a) Let p(x) and q(x) be elements of P_2 and let c be a scalar.

(i) T(c(p(x))) = T((cp(x)))

= (cp(1),cp'(1),cp''(1))

=c(p(1),p'(1),p''(1))

=cT(p(x))

(ii) T((p(x)) + (q(x)))

= T((p(x) + q(x)))

=((p+q)(1),(p+q)'(1),(p+q)''(1))

=(p(1) + q(1), p'(1) + q'(1), p''(1) + q''(1))

=(p(1),p'(1),p''(1)) + (q(1), q'(1), q''(1))

=T(p(x)) + T(q(x))

b) I don't know how to do this question. I've read through the relevant section of my textbook and there is a part which mentions a linear transformation T: R^n -> R^m and a "unique matrix" A such that Ax = T(x) for all x in R^n. I'm not sure if this is related to the question though since I am not familiar with the terminology of this question so I'm making assumptions as to what they mean - specifically "matrix of T".

Well I guess I'll just list some thoughts. If I use the Ax = T(x) method I would deduce that T is a 3 x 3 matrix. I do this by using coordinates vectors with the basis {1,x,x^2} although it appears to be a rather weak 'guess method.'

I can't think of any other methods that I can use to do this question. I'm going to think about it a bit more but could someone please help me out?