Let \(F\) be a subfield of the complex numbers and let \(T\), \(D\) be the transformations on \(F[x]\) defined by \[ T\left(\sum_{i=0}^{n} c_i x^i\right) = \sum_{i=0}^{n} \frac{c_i}{1+i} x^{i+1} \] and \[ D\left(\sum_{i=0}^{n} c_i x^i\right) = \sum_{i=1}^{n} i c_i x^{i-1}. \]
Show that \(T\) is a non-singular linear operator on \(F[x]\). Show also that \(T\) is not invertible.
Show that \(D\) is a linear operator on \(F[x]\) and find its null space.
Show that \(DT = I\), and \(TD \neq I\).
Show that \(T[(Tf)g] = (Tf)(Tg) - T[f(Tg)]\) for all \(f, g\) in \(F[x]\).
State and prove a rule for \(D\) similar to the one given for \(T\) in (d).
Suppose \(V\) is a non-zero subspace of \(F[x]\) such that \(Tf\) belongs to \(V\) for each \(f\) in \(V\). Show that \(V\) is not finite-dimensional.
Suppose \(V\) is a finite-dimensional subspace of \(F[x]\). Prove there is an integer \(m \geq 0\) such that \(D^m f = 0\) for each \(f\) in \(V\).
Solution
1. Part (a): \(T\) is non-singular but not invertible:
1.1. If \(T(f) = 0\) for \(f = \sum_{i=0}^n c_i x^i\), then \(\sum_{i=0}^n \frac{c_i}{1+i} x^{i+1} = 0\).Equating polynomial coefficients yields \(\frac{c_i}{1+i} = 0 \implies c_i = 0\) for all \(i\), so \(f = 0\).Thus \(\ker(T) = \{0\}\), which means \(T\) is non-singular (injective).
uniqueness of polynomial coefficients.
1.2. For every \(f \in F[x]\), the polynomial \(T(f) = \sum_{i=0}^n \frac{c_i}{1+i} x^{i+1}\) has no constant term (that is, \((Tf)(0) = 0\)).Thus the constant polynomial \(1\) is not in the image \(\operatorname{im}(T)\).Therefore \(T\) is not surjective, hence not invertible.
\(T(f)(0) = 0 \neq 1\).
2. Part (b): Linearity and null space of \(D\):
2.1. For \(f = \sum a_i x^i\), \(g = \sum b_i x^i\), and \(c \in F\):\[ D(cf + g) = \sum_{i=1}^n i(ca_i + b_i)x^{i-1} = c \sum_{i=1}^n i a_i x^{i-1} + \sum_{i=1}^n i b_i x^{i-1} = c D(f) + D(g). \]Thus \(D\) is linear.
definition of formal derivative.
2.2. If \(D(f) = 0\), then \(\sum_{i=1}^n i c_i x^{i-1} = 0 \implies i c_i = 0\) for all \(i \ge 1\).Since \(F\) is a subfield of \(\mathbb{C}\) (characteristic 0), \(c_i = 0\) for all \(i \ge 1\).Thus \(f(x) = c_0\) is a constant polynomial.Therefore \(\ker(D) = F\) (the subspace of constant polynomials).
characteristic 0 implies \(i \neq 0\) for \(i \ge 1\).
3. Part (c): Compositions \(DT\) and \(TD\):
3.1. For any monomial \(x^k\) (\(k \ge 0\)):\[ DT(x^k) = D\left(\frac{x^{k+1}}{k+1}\right) = (k+1) \frac{x^k}{k+1} = x^k. \]By linearity, \(DT(f) = f\) for all \(f \in F[x]\), so \(DT = I\).
evaluation on basis \(\{x^k\}\).
3.2. For the constant polynomial \(1\): \(TD(1) = T(0) = 0 \neq 1\).Thus \(TD \neq I\) (in general, \(TD(f) = f(x) - f(0)\)).
\(TD(1) = 0 \neq 1\).
4. Part (d): Integration by parts formula for \(T\):
4.1. Note that \((Tf)(0) = 0\) and \((Tg)(0) = 0\), so the product \(((Tf)(Tg))(0) = 0\).
\(Tf\) has zero constant term.
4.2. Apply the product rule for \(D\) to \((Tf)(Tg)\):\[ D\big((Tf)(Tg)\big) = \big(D(Tf)\big)(Tg) + (Tf)\big(D(Tg)\big) = f(Tg) + (Tf)g, \]since \(DT = I\) by Part (c).
derivative product rule and 3.
4.3. Applying \(T\) to both sides and using \(TD(h) = h(x) - h(0)\) with \(h = (Tf)(Tg)\) (where \(h(0) = 0\)):\[ (Tf)(Tg) = T\big(f(Tg)\big) + T\big((Tf)g\big). \]
\(TD(h) = h\) when \(h(0) = 0\).
4.4. Rearranging terms gives \(T[(Tf)g] = (Tf)(Tg) - T[f(Tg)]\).
algebraic rearrangement.
5. Part (e): Product rule for \(D\):
5.1. Rule: \(D(fg) = (Df)g + f(Dg)\) for all \(f, g \in F[x]\).
statement of Leibniz rule.
2.
On basis monomials \(f = x^i, g = x^j\):
\[ D(x^i x^j) = D(x^{i+j}) = (i+j)x^{i+j-1} = (i x^{i-1})x^j + x^i (j x^{j-1}) = (Dx^i)x^j + x^i(Dx^j). \] By bilinearity, the identity holds for all polynomials \(f, g \in F[x]\).
bilinearity of multiplication and linearity of \(D\).
6. Part (f): \(T\)-invariant non-zero subspace is infinite-dimensional:
6.1. Let \(V \neq \{0\}\) be a \(T\)-invariant subspace, and choose a non-zero \(f \in V\) of degree \(d = \deg(f) \ge 0\).
\(V\) is non-zero.
6.2. Since \(V\) is \(T\)-invariant, \(T^k(f) \in V\) for all \(k \ge 0\).
mathematical induction on \(T\)-invariance.
6.3. Since \(\deg(T(h)) = \deg(h) + 1\) for any non-zero polynomial \(h\), we have \(\deg(T^k(f)) = d + k\).
definition of \(T\).
6.4. The set of polynomials \(\{f, T(f), T^2(f), \dots, T^{m-1}(f)\} \subset V\) have strictly increasing degrees \(d, d+1, \dots, d+m-1\), hence are linearly independent for every \(m \ge 1\).
polynomials of distinct degrees are linearly independent.
6.5. Thus \(\dim(V) \ge m\) for all \(m \in \mathbb{N}\), which proves \(\dim(V) = \infty\).
existence of arbitrarily large linearly independent sets.
7. Part (g): Nilpotency of \(D\) on finite-dimensional subspaces:
7.1. Let \(V\) be a finite-dimensional subspace of \(F[x]\), and let \(\{v_1, \dots, v_k\}\) be a basis for \(V\).
finite-dimensionality.
7.2. Let \(m = \max_{1 \le j \le k} \deg(v_j) + 1\).
maximum of finitely many integers is well-defined.
7.3. Since \(\deg(D(h)) = \deg(h) - 1\), \(D^{\deg(h)+1}(h) = 0\) for every non-zero polynomial \(h\).Thus \(D^m(v_j) = 0\) for all \(j = 1, \dots, k\).
formal derivative decreases degree by 1.
7.4. For any \(f = \sum_{j=1}^k c_j v_j \in V\), linearity implies \(D^m(f) = \sum_{j=1}^k c_j D^m(v_j) = 0\).
linearity of \(D^m\).
8. Conclusion:All statements (a) through (g) are proved. Q.E.D.
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