Can i get help with this algebra 2 problem? Let P_n be the vector space of all p
ID: 3031253 • Letter: C
Question
Can i get help with this algebra 2 problem?
Let P_n be the vector space of all polynomials of degree n or less in the variable x. Let D^2: P_4 rightarrow P_2 be the linear transformation that takes a polynomial to its second derivative. That is, D^2(p(x)) = p"(x) for any polynomial p(x) of degree 4 or less. A basis for the kernel of D^2 is {}. Enter a polynomial or a comma separated list of polynomials. A basis for the image of D^2 is {}. Enter a polynomial or a comma separated list of polynomials.Explanation / Answer
An arbitrary element of P4(x) is ax4 +bx3 +cx2 +dx +e = u(say), where a, b, c, d , e are arbitrary constants. The first derivative of ax4 +bx3 +cx2 +dx +e is 4ax3 +3bx2 + 2cx +d and the second derivative of ax4 +bx3 +cx2 +dx +e is 12ax +6bx +2c = v(say). Then, the image of the linear transformation D2 i. e. Im( D2 ) = { D2(u): u P4} = {v P2(x) : v = D2 ( u), for some u P4 (x) }. Since v = 12ax +6bx +2c is the image of u = ax4 +bx3 +cx2 +dx +e under the linear transformation D2 , therefore ax +bx +c P2(x) is the image of (a/12)x4 +(b/6)x3 +(c/2)x2 +dx + e P4(x). Hence a basis for Im( D2 ) is { 1,x,x2}.
The kernel of the linear transformation D2 is the set of all vectors u P4 (x) such that D2(u) = 0 Now, as above, if u = ax4 +bx3 +cx2 +dx +e , then D2(u) = 12ax +6bx +2c . Thus, if D2(u) = 0, then a = 0, b =0 and c = 0. Then u = dx +e . Thus, a basis for the Kernel of D2 i.e. Ker (D2) = { 1,x}.