[proofplan] Part (1) is an element chase: $(U_1 + U_2)^0 = U_1^0 \cap U_2^0$. Part (2) uses the containment $U_1^0 + U_2^0 \subseteq (U_1 \cap U_2)^0$ together with a dimension count using part (1) and the [Dimension of Annihilator](/theorems/420) to prove equality. [/proofplan] [step:Prove $(U_1 + U_2)^0 = U_1^0 \cap U_2^0$ by element chase] ($\subseteq$): If $\theta \in (U_1 + U_2)^0$, then $\theta$ vanishes on $U_1 + U_2$. Since $U_1 \subseteq U_1 + U_2$ and $U_2 \subseteq U_1 + U_2$, the functional $\theta$ vanishes on both $U_1$ and $U_2$. So $\theta \in U_1^0 \cap U_2^0$. ($\supseteq$): If $\theta \in U_1^0 \cap U_2^0$, then $\theta$ vanishes on $U_1$ and on $U_2$. For any $u_1 + u_2 \in U_1 + U_2$: \begin{align*} \theta(u_1 + u_2) = \theta(u_1) + \theta(u_2) = 0 + 0 = 0. \end{align*} So $\theta \in (U_1 + U_2)^0$. [/step] [step:Prove $(U_1 \cap U_2)^0 = U_1^0 + U_2^0$ by containment and dimension counting] **Containment $U_1^0 + U_2^0 \subseteq (U_1 \cap U_2)^0$:** For $\theta \in U_1^0$ and $\phi \in U_2^0$ and any $v \in U_1 \cap U_2$: \begin{align*} (\theta + \phi)(v) = \theta(v) + \phi(v) = 0 + 0 = 0, \end{align*} since $v \in U_1$ (so $\theta(v) = 0$) and $v \in U_2$ (so $\phi(v) = 0$). **Dimension count:** By [Dimension of Annihilator](/theorems/420): \begin{align*} \dim(U_1 \cap U_2)^0 = \dim V - \dim(U_1 \cap U_2). \end{align*} By the [Dimension Formula for Sums and Intersections](/theorems/374): $\dim(U_1 \cap U_2) = \dim U_1 + \dim U_2 - \dim(U_1 + U_2)$. Using part (1) and the same dimension formula in $V^*$: \begin{align*} \dim(U_1^0 + U_2^0) &= \dim U_1^0 + \dim U_2^0 - \dim(U_1^0 \cap U_2^0) \\ &= (\dim V - \dim U_1) + (\dim V - \dim U_2) - (\dim V - \dim(U_1 + U_2)) \\ &= \dim V - \dim U_1 - \dim U_2 + \dim(U_1 + U_2) \\ &= \dim V - \dim(U_1 \cap U_2). \end{align*} Since $U_1^0 + U_2^0 \subseteq (U_1 \cap U_2)^0$ and both have the same dimension, they are equal. [/step]