Define T P3 P3 by Tpxpx px is the derivative of px a show t

Define T: P3 --> P3 by T(p)=xp\'(x) (p\'(x) is the derivative of p(x)

a) show that T is a linear operator

b) find image of p(x) = ax^2 + bx + c under T. (a,b,c are constants)

c) is T 1-1 mapping

d) Is T onto?

Solution

a)

T(0)=0

T(p+q)=x(p+q)\'(x)=xp\'(x)+xq\'(x)=T(p)+T(q)

T(cp)=x(cp)\'=cxp\'=cT(p)

Hence, T is linear.

(b)

T(ax^2+bx+c)=x(ax^2+bx+c)\'=x(2ax+b)=2ax^2+bx

(c)

No it is not:

T(x^2+x+1)=T(x^2+x)

Because image of T depends only on constants a,b and not c. So we can vary c arbitrarily without changing a,b which will leave the image unchanged.

d)

T(ax^2+bx+c)=2ax^2+bx

Hence constants polynomials are not present in image of T

So T is not onto.