Question

Find a basis for the subspace of Pn defined by V={p an element of Pn, such that p(1)=0}. What is the dimension of V?

Answer 1

Let p(x) = a_nxⁿ+ a_n-1x^n-1+…+a₁x + a₀ be an arbitrary polynomial in P_n. If p(1) = 0, then we have a_n+a_n-1+…+a₀ = 0.

Now, let p(x)= a_nxⁿ+ a_n-1x^n-1+…+a₁x + a₀ and q(x) = b_nxⁿ+ b_n-1x^n-1+…+b₁x + b₀ be 2 arbitrary polynomial in V and let k be an arbitrary scalar. Then a_n+a_n-1+…+a₀ = 0 and b_n+b_n-1+…+b₀ = 0. Further, p(x)+q(x) = a_nxⁿ+ a_n-1x^n-1+…+a₁x + a₀ + b_nxⁿ+ b_n-1x^n-1+…+b₁x + b₀ = (a_n+b_n)xⁿ+(a_n-1+b_n-1)x^n-1 +…+(a₁+b₁)x+(a₀+b₀). Also, (a_n+b_n)+(a_n-1+b_n-1) +…+(a₁+b₁)+(a₀+b₀)= (a_n+a_n-1+…+a₀ )+ (b_n+b_n-1+…+b₀ )=0+0= 0. This implies that p(x)+q(x) ∈ V so that V is closed under vector addition. Also, kp(x) = k(a_nxⁿ+ a_n-1x^n-1+…+a₁x + a₀) = ka_nxⁿ+ ka_n-1x^n-1+…+ka₁x + ka₀. Now, since ka_n+ka_n-1+…+ak₀ =k(a_n+a_n-1+…+a₀ )= k*0 = 0, hence kp(x) ∈ V so that V is closed under scalar multiplication. Further, apparently, the zero polynomial belongs to V as p(x) = 0 when all the a_i s are 0. Hence V is a vector space and , therefore, a subspace of P_n .

Now, if a_n+a_n-1+…+a₁+a₀ = 0, then a₀ = -( a_n+a_n-1+…+a₁) so that p(x) = a_nxⁿ+ a_n-1x^n-1+…+a₁x -( a_n+a_n-1+…+a₁) = a_n(xⁿ-1)+ a_n-1 (x^n-1 -1)+…+a₁ (x-1). This implies that every polynomial in V is a linear combination of (xⁿ-1), (x^n-1 -1), (x-1) which are apparently linearly independent. Hence the set{(xⁿ-1), (x^n-1 -1), (x-1) } is a basis for V. Therefore, the dimension of V is n.