Let \(F\) be a subfield of the field of complex numbers and let \(V\) be any vector space over \(F\). Suppose that \(f\) and \(g\) are linear functionals on \(V\) such that the function \(h\) defined by \(h(\alpha) = f(\alpha)g(\alpha)\) is also a linear functional on \(V\). Prove that either \(f = 0\) or \(g = 0\).
Solution
1. Suppose for contradiction that \(f \neq 0\) and \(g \neq 0\).
assume both are nonzero.
2. Since \(h\) is linear, \(h(c\alpha) = c\,h(\alpha)\) for every scalar \(c\) and every \(\alpha \in V\).
linearity of \(h\).
3. Expanding: \(f(c\alpha)g(c\alpha) = c f(\alpha)g(\alpha)\), i.e. \(c^2 f(\alpha)g(\alpha) = c f(\alpha)g(\alpha)\).
2 and linearity of \(f, g\).
4. Hence \((c^2 - c) f(\alpha)g(\alpha) = 0\) for all scalars \(c\) and all \(\alpha \in V\).
3.
5. Taking \(c = 2\) (so \(c^2 - c = 2 \neq 0\)), we get \(f(\alpha)g(\alpha) = 0\) for all \(\alpha \in V\).
4.
6. Hence \(h \equiv 0\), i.e. \(f(\alpha)g(\alpha) = 0\) for every \(\alpha\).
5.
7. Since \(f \neq 0\), there is \(\alpha_0\) with \(f(\alpha_0) \neq 0\); then \(g(\alpha_0) = 0\) by
6.
1 and 6.
8. For any \(\beta \in V\), \(f(\alpha_0 + \beta)g(\alpha_0 + \beta) = 0\) by
6.
6 applied to \(\alpha_0 + \beta\).
9. Expanding: \((f(\alpha_0) + f(\beta))(g(\alpha_0) + g(\beta)) = 0\), and since \(g(\alpha_0) = 0\), this gives \((f(\alpha_0) + f(\beta))g(\beta) = 0\).
7 and 8.
10. Hence \(f(\alpha_0)g(\beta) + f(\beta)g(\beta) = 0\), and \(f(\beta)g(\beta) = 0\) by 6, so \(f(\alpha_0)g(\beta) = 0\).
9 and 6.
11. Since \(f(\alpha_0) \neq 0\), this forces \(g(\beta) = 0\) for every \(\beta \in V\).
10.
12. Hence \(g = 0\), contradicting
1.
11.
13. Therefore \(f = 0\) or \(g = 0\).
12.
14. Q.E.D.
13.
Solution
Since \(h\) is linear, for every \(\alpha\in V\), \[ h(2\alpha)=2h(\alpha). \] But by its definition, \[ h(2\alpha)=f(2\alpha)g(2\alpha)=4f(\alpha)g(\alpha)=4h(\alpha). \] Because \(F\) is a subfield of \(\CC\), its characteristic is zero; hence \[ h(\alpha)=0 \] for every \(\alpha\). Thus \[ f(\alpha)g(\alpha)=0\qquad\text{for all }\alpha\in V. \] Suppose both \(f\) and \(g\) are nonzero. Choose \(u,v\in V\) with \(f(u)\ne0\) and \(g(v)\ne0\). The displayed identity forces \(g(u)=0\) and \(f(v)=0\). Then \[ f(u+v)g(u+v)=f(u)g(v)\ne0, \] a contradiction. Therefore either \(f=0\) or \(g=0\).