Let V = P_5, the vector space of polynomials of degree lessthanorequalto 5, with

Question

Let V = P_5, the vector space of polynomials of degree lessthanorequalto 5, with coefficients in R, and let W = {p(x) element P_5| p(0) = p(1) = p(2)} Show that W is a subspace of V. Verify that p(x) = 2x^5 - 4x^4 + 2x^3 - 8x^2 + 8x + 7 belongs to W. For p as in part (b), express p(x) = x(x - 1)(x - 2) r(x) + p(0), with r of degrees lessthanorequalto 2. Let u(x) = x(x - 1)(x - 2), and explain why for every q element W there exists r element P_2 such that q(x) = u(x) middot r(x) + q(0). Find a basis B for W and order it by increasing degree. Find the dimension of W. For u as in part (d), find the coordinate vector [u]_B of u with respect to B from part (e).

Explanation / Answer

V is the vector space = P5= polynomial of degree <= 5

W = { P(x) / P(0) = P(1)= P (2) } is a subset as it is an element of V satisfying certain extra conditions

to veify whether it is a subspace or not , let us use the necessary and sufficient cond n for a subspace

Let P1(x) , P2(x) are in W

P1(0) = P1(1) = P1(2) and P2(0) = P2(1) = P 2(2) -----(1)

consider P(x)= [ c1P1+c2P2](x)

P (0) =[c1P1+c2P2] (0)= c1P1(0) +c2P2(0) = P(1) =P(2) from (1)

Hence P(x) is in W => W is the subspace of "V'

b. P(x) = 2x5- 4x4+2x3-8x2+8x +7

P(0) =7 , P(1) = 2-4-2-8+8+7=7 , P(2) = 64-64+16-32+16+7=7

P(0)=P(1) = P(2)=7

c . P(x) = x(x-1) (x-2) r(x) +7 equating the given P(x)

we get r(x) = 2x2+2x+4

d. U(x) =x(x-1)(x-2)

Q(x) = U(x) r(x) + Q(0)

Q(0)= Q(0) , Q(1) = Q(0) , Q(2) =Q(0) { as U(0) =0 , U(1) =0 , U(2) =0 }

hence Q(x) is an element of W

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