linear algebra p0 1 Define a Linear Transformation T Ps R3 a

linear algebra

p(0) 1) Define a Linear Transformation T: Ps R3 as T (p) SpK0) p00) li) Find a basis of kernel of T Find a basis for range of T Note: Any element of Ps is of the fo where the coefficients a,b, c, d,e,f are real numbers. Hint: Any set of functions fi f (t) fa(e) where f t\", f2()- t a..... t\" where n n2, ns, are distinct integers are linearly independent

Solution

An arbitrary element of P₅ is of the form ax⁵ +bx⁴ +cx³ +dx² +ex + f where a, b, c, d, e, f R, the set of real numbers. The kernel of the linear transformation T, i.e. Ker (T) is the set of all the polynomials p P₅ such T(p) = 0. Further, since T(p) = [ p(0), 5p(0), p(0) ]^T , therefore for all polynomials p in Ker (T), we have [ p(0), 5p(0), p(0) ]^T = [ 0, 0, 0]^T   so that p(0) = 0 .Since p(0) = f , the constant term in the polynomial p = ax⁵ +bx⁴ +cx³ +dx² +ex + f , therefore, f = 0. Thus, Ker(T) = { p = ax⁵ +bx⁴ +cx³ +dx² +ex : a, b, c, d, e, f R, } i.e. Ker(T) is the set of all polynomials in P₅ which do not have any constant term. Thus, a basis for Ker(T) is { x⁵, x⁴, x³, x², x}

The range of T is the set of all the vectors v in R³ such that there is a polynomial p in P₅ with T(p) =  v . However, T(p) = [ p(0), 5p(0), p(0) ]^T so that v = [ p(0), 5p(0), p(0) ]^T . Further, as deduced above, p(0) = f , the constant term in the polynomial p=ax⁵ +bx⁴ +cx³ +dx² +ex + f . Thus, the range of T is the set of all vectors of the form [ f, 5f, f ]^T where f is a real number. Thus, a basis for the Range of T is { [1,5,1]^T}