Let \(n\) be a positive integer and let \(V\) be the space of all polynomial functions over the field of real numbers which have degree at most \(n\), i.e., functions of the form \[ f(x) = c_0 + c_1 x + \cdots + c_n x^n. \] Let \(D\) be the differentiation operator on \(V\). Find a basis for the null space of the transpose operator \(D^t\).
Solution
Goal. For \(V\) the space of polynomials of degree \(\le n\) and \(D\) differentiation, find a basis for \(\ker D^t\).
1. \(D^t: V^* \to V^*\) is defined by \((D^t \ell)(f) = \ell(Df)\).
definition of the transpose (dual) map.
2. \(\ker D^t = \theset{\ell \in V^* : \ell(Df) = 0 \text{ for all } f \in V}\).
\(\ell \in \ker D^t\) iff \(D^t \ell = 0\) iff \(\ell(Df) = 0\) for all \(f\).
3. \(\im D = \theset{\text{polynomials of degree } \le n-1}\).
\(D\) maps \(V\) onto the polynomials of degree \(\le n-1\) (the derivative of a degree-\(\le n\) polynomial has degree \(\le n-1\), and every such is a derivative).
4. \(\ker D^t = (\im D)^\perp = \theset{\ell \in V^* : \ell|_{\im D} = 0}\).
\(\ell(Df) = 0\) for all \(f\) iff \(\ell\) vanishes on \(\im D\).
5. \(\im D\) has codimension \(1\) in \(V\) (it is the subspace of polynomials of degree \(\le n-1\)).
\(\dim V = n+1\) and \(\dim \im D = n\).
6. Hence \(\ker D^t\) is \(1\)-dimensional.
\(\dim \ker D^t = \dim V - \dim \im D = (n+1) - n = 1\).
7. A basis for \(\ker D^t\) is \(\theset{\ell}\) where \(\ell(f) = f^{(n)}(0)\) (the \(n\)-th derivative evaluated at \(0\)), or equivalently the functional \(\ell(f) = c_n\) (the leading coefficient).
\(\ell(Df) = (Df)^{(n)}(0) = f^{(n+1)}(0) = 0\) for all \(f\) (since \(f\) has degree \(\le n\)), so \(\ell \in \ker D^t\); and \(\ell \neq 0\) (e.g. \(\ell(x^n) = n! \neq 0\)), so it spans the \(1\)-dimensional kernel.
8. Q.E.D.
7 gives the basis \(\theset{\ell}\) with \(\ell(f) = c_n\) (the coefficient of \(x^n\)).