Let \(V\) be the inner product space of Exercise 8, and let \(D\) be the differentiation operator on \(V\). Find \(D^*\).
Solution
Exercise 8 uses \[ V=\{a_0+a_1x+a_2x^2+a_3x^3:a_i\in\mathbb R\} \] with \[ (f\mid g)=\int_0^1 f(t)g(t)\,dt. \] Let \[ \mathfrak B=(1,x,x^2,x^3). \]
1. Write the Gram matrix and the matrix of differentiation in \(\mathfrak B\).
The Gram matrix is the \(4\times4\) Hilbert matrix \[ G=\left(\frac1{i+j+1}\right)_{0\le i,j\le3} = \begin{pmatrix} 1&1/2&1/3&1/4\\ 1/2&1/3&1/4&1/5\\ 1/3&1/4&1/5&1/6\\ 1/4&1/5&1/6&1/7 \end{pmatrix}. \] Since \[ D(1)=0,\quad D(x)=1,\quad D(x^2)=2x,\quad D(x^3)=3x^2, \] we have \[ [D]_{\mathfrak B}= \begin{pmatrix} 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\\ 0&0&0&0 \end{pmatrix}. \]
2. The adjoint matrix is \(G^{-1}[D]^tG\).
For coordinate columns \(u,v\), the inner product is \[ (u\mid v)=u^tGv. \] If \(A=[D]_{\mathfrak B}\) and \(B=[D^*]_{\mathfrak B}\), then \[ (Au)^tGv=u^tGBv \] for all \(u,v\). Hence \[ A^tG=GB, \qquad B=G^{-1}A^tG. \] Direct exact arithmetic gives \[ \boxed{ [D^*]_{\mathfrak B}= \begin{pmatrix} -20&-5&-4&-4\\ 180&60&58&60\\ -420&-180&-180&-183\\ 280&140&140&140 \end{pmatrix}.} \] This is the adjoint matrix relative to the monomial basis [@HK71].
3. Equivalently, \(D^*\) is given on the basis vectors by the four columns above.
Thus \[ \begin{aligned} D^*(1)&=-20+180x-420x^2+280x^3,\\ D^*(x)&=-5+60x-180x^2+140x^3,\\ D^*(x^2)&=-4+58x-180x^2+140x^3,\\ D^*(x^3)&=-4+60x-183x^2+140x^3. \end{aligned} \] These formulas determine \(D^*\) uniquely on all of \(V\).