[proofplan] Consider the restriction map $\rho: V^* \to U^*$ defined by $\rho(\theta) = \theta|_U$. Its kernel is $U^0$ and it is surjective (any functional on $U$ extends to $V$ by the [Basis Extension Theorem](/theorems/370)). Rank-Nullity gives $\dim V^* = \dim U^0 + \dim U^*$, and $\dim V^* = \dim V$, $\dim U^* = \dim U$ by the [Dimension of Dual Space](/theorems/415). [/proofplan] [step:Define the restriction map and identify its kernel as $U^0$] Define $\rho: V^* \to U^*$ by $\rho(\theta) = \theta|_U$ (restrict the domain of $\theta$ from $V$ to $U$). This is linear since restriction preserves addition and scalar multiplication. A functional $\theta$ lies in $\ker\rho$ iff $\theta|_U = 0$ iff $\theta(u) = 0$ for all $u \in U$ iff $\theta \in U^0$. [/step] [step:Prove surjectivity of $\rho$ by extending functionals] Let $\phi \in U^*$. Choose a basis $(u_1, \dots, u_k)$ for $U$ and extend to a basis $(u_1, \dots, u_k, v_{k+1}, \dots, v_n)$ for $V$ (by the [Basis Extension Theorem](/theorems/370)). By the [Universal Property of Linear Maps](/theorems/380), there exists $\theta \in V^*$ with $\theta(u_i) = \phi(u_i)$ for $i = 1, \dots, k$ and $\theta(v_j) = 0$ for $j = k+1, \dots, n$. Then $\rho(\theta) = \theta|_U = \phi$, so $\rho$ is surjective. [/step] [step:Apply Rank-Nullity to conclude] By [Rank-Nullity](/theorems/384) applied to $\rho: V^* \to U^*$: \begin{align*} \dim V^* = \dim\ker\rho + \dim\mathrm{im}\,\rho = \dim U^0 + \dim U^*. \end{align*} By [Dimension of Dual Space](/theorems/415), $\dim V^* = \dim V$ and $\dim U^* = \dim U$. Therefore $\dim V = \dim U^0 + \dim U$, giving $\dim U^0 = \dim V - \dim U$. [/step]