Let \(V\) be the subspace of \(R[x]\) of polynomials of degree at most 3. Equip \(V\) with the inner product \[ (f \mid g) = \int_0^1 f(t)g(t)\,dt. \]
Find the orthogonal complement of the subspace of scalar polynomials.
Apply the Gram-Schmidt process to the basis \(\{1, x, x^2, x^3\}\).
Solution
1. The orthogonal complement of the scalar polynomials is the zero-integral hyperplane.
The scalar polynomials form \(\operatorname{span}\{1\}\). Thus \[ f\perp1 \iff (f\mid1)=0 \iff \int_0^1f(t)\,dt=0. \] Hence \[ \boxed{ (\operatorname{span}\{1\})^\perp =\left\{a+bx+cx^2+dx^3: a+\frac b2+\frac c3+\frac d4=0\right\}.} \] For example, a convenient basis is \[ \boxed{ x-\frac12, \quad x^2-\frac13, \quad x^3-\frac14.} \] These three vectors satisfy the zero-integral condition and are linearly independent, so they form a basis of the three-dimensional hyperplane.
2. Gram–Schmidt on \((1,x,x^2,x^3)\) yields the shifted Legendre basis.
Starting with \(q_0=1\), one has \(\|q_0\|=1\). The next residual is \[ q_1=x-\frac12, \] with \[ \|q_1\|^2=\frac1{12}. \] For \(x^2\), subtracting its projections onto \(q_0\) and \(q_1\) gives \[ q_2=x^2-x+\frac16, \qquad \|q_2\|^2=\frac1{180}. \] For \(x^3\), the next Gram–Schmidt residual is \[ q_3=x^3-\frac32x^2+\frac35x-\frac1{20}, \qquad \|q_3\|^2=\frac1{2800}. \] Each norm follows by integrating the square on \([0,1]\), and direct integration shows \((q_i\mid q_j)=0\) for \(i\ne j\) [@HK71].
Normalizing gives the orthonormal basis \[ \boxed{ \begin{aligned} u_0&=1,\\ u_1&=\sqrt3(2x-1),\\ u_2&=\sqrt5(6x^2-6x+1),\\ u_3&=\sqrt7(20x^3-30x^2+12x-1). \end{aligned}} \] These are exactly the vectors obtained by Gram–Schmidt from the prescribed ordered basis.