Let \(V\) be the vector space of all polynomial functions \(p\) from \(\RR\) into \(\RR\) which have degree 2 or less: \[ p(x) = c_0 + c_1 x + c_2 x^2. \] Define three linear functionals on \(V\) by \[ f_1(p) = \int_0^1 p(x)\,dx, \quad f_2(p) = \int_0^2 p(x)\,dx, \quad f_3(p) = \int_0^{-1} p(x)\,dx. \] Show that \(\{f_1, f_2, f_3\}\) is a basis for \(V^*\) by exhibiting the basis for \(V\) of which it is the dual.
Solution
Goal. Show \(\theset{f_1, f_2, f_3}\) is a basis of \(V^*\) by finding the dual basis of \(V\).
1. Find polynomials \(p_1, p_2, p_3 \in V\) with \(f_i(p_j) = \delta_{ij}\).
1.1. Write \(p_j(x) = c_{0j} + c_{1j} x + c_{2j} x^2\) and impose \(f_i(p_j) = \delta_{ij}\).
\(f_i(p_j) = c_{0j}\int_0^{a_i} 1\,dx + c_{1j}\int_0^{a_i} x\,dx + c_{2j}\int_0^{a_i} x^2\,dx\) where \(a_1 = 1, a_2 = 2, a_3 = -1\).
1.2. Compute the integrals: \(\int_0^a 1\,dx = a\), \(\int_0^a x\,dx = a^2/2\), \(\int_0^a x^2\,dx = a^3/3\).
elementary antiderivatives.
1.3. The matrix of the system is\[ \begin{pmatrix} 1 & 1/2 & 1/3 \\ 2 & 2 & 8/3 \\ -1 & 1/2 & -1/3 \end{pmatrix} \]with rows indexed by \(f_1, f_2, f_3\) and columns by \(1, x, x^2\).
substitute \(a = 1, 2, -1\) into 2.2.
1.4. This matrix is invertible, so the system \(f_i(p_j) = \delta_{ij}\) has a unique solution for each \(j\).
the determinant is nonzero (the three functionals are linearly independent, since the rows are independent).
2. The resulting \(\theset{p_1, p_2, p_3}\) is a basis of \(V\) dual to \(\theset{f_1, f_2, f_3}\).
2.1. \(\theset{p_1, p_2, p_3}\) is linearly independent.
if \(\sum_j \lambda_j p_j = 0\), applying \(f_i\) gives \(\lambda_i = 0\) for each \(i\).
2.2. Hence \(\theset{p_1, p_2, p_3}\) is a basis of \(V\) (which has dimension \(3\)).
three linearly independent vectors in a \(3\)-dimensional space.
2.3. \(\theset{f_1, f_2, f_3}\) is the dual basis, hence a basis of \(V^*\).
\(f_i(p_j) = \delta_{ij}\) is exactly the definition of a dual basis.
3. Q.E.D.
2.3 shows \(\theset{f_1, f_2, f_3}\) is a basis of \(V^*\).
Solution
Relative to the standard basis \(\{1,x,x^2\}\), the three functionals have matrix \[ M= \begin{bmatrix} 1&\frac12&\frac13\\ 2&2&\frac83\\ -1&\frac12&-\frac13 \end{bmatrix}. \] Its determinant is \(-2\), so the functionals are linearly independent and hence form a basis of \(V^*\).
To exhibit the basis of \(V\) of which they are dual, use the columns of \(M^{-1}\): \[ M^{-1}= \begin{bmatrix} 1&-\frac16&-\frac13\\ 1&0&1\\ -\frac32&\frac12&-\frac12 \end{bmatrix}. \] Thus set \[ p_1(x)=1+x-\frac32x^2, \qquad p_2(x)=-\frac16+\frac12x^2, \qquad p_3(x)=-\frac13+x-\frac12x^2. \] The identity \(MM^{-1}=I\) gives \[ f_i(p_j)=\delta_{ij}. \] Hence \(\{p_1,p_2,p_3\}\) is a basis of \(V\) and \(\{f_1,f_2,f_3\}\) is its dual basis.