Let \(T\) be a linear operator on \(V\). If every subspace of \(V\) is invariant under \(T\), then \(T\) is a scalar multiple of the identity operator.
Solution
1. Every non-zero vector is an eigenvector of \(T\):
1.1. Let \(v \in V \setminus \{0\}\). The 1-dimensional subspace \(W = \operatorname{span}(v) = F v\) is a subspace of \(V\).By hypothesis, \(W\) is invariant under \(T\), so \(T(W) \subseteq W\).
1.2. Therefore \(T(v) \in F v\), which implies there exists a scalar \(c_v \in F\) such that:\[ T(v) = c_v v. \]
2. Proof that the scalar \(c_v\) is independent of \(v\):
2.1. If \(\dim(V) \le 1\), then every linear operator on \(V\) is a scalar multiple of the identity: when \(\dim(V) = 0\) the only operator is the zero map, and when \(\dim(V) = 1\) any operator sends a basis vector \(e\) to \(c e\) for some \(c \in F\), hence acts as \(c \operatorname{id}_V\). Assume \(\dim(V) \ge 2\).Let \(u, v \in V \setminus \{0\}\).
2.2. Subcase 1: \(\{u, v\}\) is linearly dependent.There exists a non-zero scalar \(\alpha \in F^\times\) such that \(v = \alpha u\).Then by linearity:\[ c_v v = T(v) = T(\alpha u) = \alpha T(u) = \alpha (c_u u) = c_u (\alpha u) = c_u v. \]Since \(v \neq 0\), \(c_v = c_u\).
2.3. Subcase 2: \(\{u, v\}\) is linearly independent.Consider the sum vector \(w = u + v \neq 0\).By 1, there exists \(c_w \in F\) such that \(T(w) = c_w w = c_w u + c_w v\).By linearity of \(T\):\[ T(w) = T(u + v) = T(u) + T(v) = c_u u + c_v v. \]Equating both expressions gives:\[ (c_u - c_w) u + (c_v - c_w) v = 0. \]Because \(\{u, v\}\) is linearly independent, \(c_u - c_w = 0\) and \(c_v - c_w = 0\), which yields \(c_u = c_v = c_w\).
3. Conclusion:Fix a non-zero vector \(v_0 \in V\) and set \(c = c_{v_0} \in F\).Then \(T(v) = c v\) for all \(v \in V \setminus \{0\}\), and for \(v = 0\), \(T(0) = 0 = c \cdot 0\).Thus \(T = c \operatorname{id}_V\). Q.E.D.