Let \(V\) be a finite-dimensional vector space over the field \(F\) and let \(T\) be a linear operator on \(V\). Let \(c\) be a scalar and suppose there is a non-zero vector \(\alpha\) in \(V\) such that \(T\alpha = c\alpha\). Prove that there is a non-zero linear functional \(f\) on \(V\) such that \(T^t f = cf\).
Solution
Goal: Prove that if \(c\) is an eigenvalue of \(T\) with nonzero eigenvector \(\alpha \in V\), then \(c\) is an eigenvalue of the transpose operator \(T^t\) with a nonzero eigenvector (linear functional) \(f \in V^*\).
1. The operator \(S = T - cI\) has a proper image:
1.1. Define the linear operator \(S = T - cI\) on \(V\).
1.2. By hypothesis, \(S\alpha = (T - cI)\alpha = T\alpha - c\alpha = 0\) with \(\alpha \neq 0\), so \(\alpha \in \ker(S)\).
1.3. Thus \(\dim \ker(S) \ge 1\).
1.4. By the Rank-Nullity Theorem for finite-dimensional \(V\):\[\operatorname{rank}(S) = \dim \operatorname{im}(S) = \dim V - \dim \ker(S) \le \dim V - 1 < \dim V.\]
1.5. Therefore, the image \(W = \operatorname{im}(S) = \{Sv : v \in V\}\) is a proper subspace of \(V\).
2. Duality relation between the kernel of \(S^t\) and the annihilator of \(\operatorname{im}(S)\):
2.1. For any linear functional \(f \in V^*\), the transpose map is defined by \((S^t f)(v) = f(Sv)\) for all \(v \in V\).
2.2. A functional \(f \in \ker(S^t)\) if and only if \(S^t f = 0\) in \(V^*\).
2.3. \(S^t f = 0 \iff (S^t f)(v) = 0 \text{ for all } v \in V \iff f(Sv) = 0 \text{ for all } v \in V \iff f(w) = 0 \text{ for all } w \in \operatorname{im}(S)\).
2.4. By definition of the annihilator subspace \((\operatorname{im} S)^0 \subset V^*\), this proves\[\ker(S^t) = (\operatorname{im} S)^0.\]
3. Existence of a nonzero annihilating functional:
3.1. For any subspace \(W \subseteq V\) of a finite-dimensional space \(V\), the dimension of its annihilator is \(\dim W^0 = \dim V - \dim W\).
3.2. Applying this to \(W = \operatorname{im}(S)\):\[\dim \ker(S^t) = \dim (\operatorname{im} S)^0 = \dim V - \operatorname{rank}(S) = \dim \ker(S) \ge 1.\]
3.3. Since \(\dim \ker(S^t) \ge 1\), there exists a nonzero linear functional \(f \in V^*\) such that \(S^t f = 0\).
4. Conclusion:
4.1. Since \(S = T - cI\), the transpose operator satisfies \(S^t = (T - cI)^t = T^t - c I_{V^*}\).
4.2. Then \(S^t f = 0\) is equivalent to \((T^t - c I_{V^*}) f = 0\), which means \(T^t f = c f\).
4.3. Since \(f \neq 0\), \(f\) is the required nonzero characteristic functional for \(T^t\) corresponding to eigenvalue \(c\).