Let \(V\) be a complex vector space and let \(f\) be a (sesqui-linear) form on \(V\) which is symmetric: \(f(\alpha, \beta) = f(\beta, \alpha)\). What is \(f\)?
Solution
1. Definition of sesquilinear form:
1.1. By convention (Hoffman–Kunze), a sesquilinear form \(f: V \times V \to \mathbb{C}\) on a complex vector space \(V\) is linear in the first argument and conjugate-linear in the second argument:\[ f(c\alpha, \beta) = c f(\alpha, \beta), \quad f(\alpha, c\beta) = \overline{c} f(\alpha, \beta) \quad \text{for all } c \in \mathbb{C}, \, \alpha, \beta \in V. \]
definition of sesquilinear form.
2. Combine symmetry with scalar multiplication:
2.1. Assume \(f\) is symmetric: \(f(\alpha, \beta) = f(\beta, \alpha)\) for all \(\alpha, \beta \in V\).
hypothesis.
2.2. For any \(c \in \mathbb{C}\) and any \(\alpha, \beta \in V\), compute \(f(c\alpha, \beta)\) in two ways:- By linearity in the first argument:\[ f(c\alpha, \beta) = c f(\alpha, \beta). \]- By symmetry and conjugate-linearity in the second argument:\[ f(c\alpha, \beta) = f(\beta, c\alpha) = \overline{c} f(\beta, \alpha) = \overline{c} f(\alpha, \beta). \]
1 and 1.
2.3. Equating both expressions gives:\[ (c - \overline{c}) f(\alpha, \beta) = 0 \quad \text{for all } c \in \mathbb{C}, \, \alpha, \beta \in V. \]
subtraction of equal quantities.
3. Deduce that \(f\) is identically zero:
3.1. Choose \(c = i \in \mathbb{C}\).Then \(c - \overline{c} = i - (-i) = 2i \neq 0\).
properties of the imaginary unit \(i\).
3.2. Substituting \(c = i\) into 3 yields:\[ 2i f(\alpha, \beta) = 0 \implies f(\alpha, \beta) = 0. \]Since this holds for all \(\alpha, \beta \in V\), \(f \equiv 0\) is the zero form.
division by \(2i \neq 0\).
4. Conclusion:\(f\) is the identically zero form (\(f = 0\)). Q.E.D.
2 and 3.