[proofplan] For part (1), we show $U \subseteq U^{00}$ directly by checking that $\mathrm{ev}(u)$ annihilates every element of $U^0$, then conclude equality by the Dimension of Annihilator theorem applied twice. For part (2), we verify $\alpha^{**} \circ \mathrm{ev}_V = \mathrm{ev}_W \circ \alpha$ by evaluating both sides at arbitrary vectors and functionals, using the definitions of the dual map and the evaluation map. [/proofplan] [step:Show $U \subseteq U^{00}$ under the canonical identification $V \cong V^{**}$] For any $u \in U$ and $\theta \in U^0$: \begin{align*} \mathrm{ev}(u)(\theta) = \theta(u) = 0, \end{align*} since $\theta \in U^0$ means $\theta$ vanishes on $U$. So $\mathrm{ev}(u) \in (U^0)^0 = U^{00}$. Under the identification $V \cong V^{**}$ via $\mathrm{ev}$, this gives $U \subseteq U^{00}$. [/step] [step:Conclude $U^{00} = U$ by dimension counting] By the [Dimension of Annihilator](/theorems/420) applied to $U \subseteq V$: \begin{align*} \dim U^0 = \dim V - \dim U. \end{align*} Applying the same theorem to $U^0 \subseteq V^*$: \begin{align*} \dim U^{00} = \dim V^* - \dim U^0 = \dim V - (\dim V - \dim U) = \dim U. \end{align*} Since $U \subseteq U^{00}$ and $\dim U = \dim U^{00}$, we have $U = U^{00}$. [guided] The containment $U \subseteq U^{00}$ was established in the previous step. To upgrade this to equality, we compare dimensions. The [Dimension of Annihilator](/theorems/420) states that for a subspace $W$ of a finite-dimensional space $X$, $\dim W^0 = \dim X - \dim W$. Applying this to $U \subseteq V$: \begin{align*} \dim U^0 = \dim V - \dim U. \end{align*} Now $U^0$ is a subspace of $V^*$, and $U^{00} = (U^0)^0$ is taken inside $V^{**}$. Applying the dimension formula again to $U^0 \subseteq V^*$: \begin{align*} \dim U^{00} = \dim V^* - \dim U^0 = \dim V - (\dim V - \dim U) = \dim U. \end{align*} We used $\dim V^* = \dim V$ from the [Dimension of Dual Space](/theorems/415). Since $U \subseteq U^{00}$ and both have dimension $\dim U$, equality holds: $U = U^{00}$. The double annihilator thus acts as an involution on the lattice of subspaces of $V$ (in finite dimensions). This is the algebraic counterpart of the fact that in a Hilbert space, $W^{\perp\perp} = \overline{W}$ (the double orthogonal complement recovers the closure, which for finite-dimensional subspaces is the subspace itself). [/guided] [/step] [step:Verify $\alpha^{**} = \alpha$ under the canonical identifications] Under the identifications $\mathrm{ev}_V: V \xrightarrow{\sim} V^{**}$ and $\mathrm{ev}_W: W \xrightarrow{\sim} W^{**}$, we must show $\alpha^{**} \circ \mathrm{ev}_V = \mathrm{ev}_W \circ \alpha$. For any $v \in V$ and $\theta \in W^*$: \begin{align*} (\alpha^{**}(\mathrm{ev}_V(v)))(\theta) &= \mathrm{ev}_V(v)(\alpha^*(\theta)) = \alpha^*(\theta)(v) = \theta(\alpha(v)) = \mathrm{ev}_W(\alpha(v))(\theta). \end{align*} Since this holds for all $\theta \in W^*$, we have $\alpha^{**}(\mathrm{ev}_V(v)) = \mathrm{ev}_W(\alpha(v))$ for all $v \in V$. That is, $\alpha^{**} \circ \mathrm{ev}_V = \mathrm{ev}_W \circ \alpha$. Identifying $V$ with $V^{**}$ and $W$ with $W^{**}$ via these evaluation maps, $\alpha^{**} = \alpha$. [guided] The claim $\alpha^{**} = \alpha$ should be understood as a commutativity statement for the diagram \begin{align*} V \xrightarrow{\alpha} W \xrightarrow{\mathrm{ev}_W} W^{**}, \qquad V \xrightarrow{\mathrm{ev}_V} V^{**} \xrightarrow{\alpha^{**}} W^{**}. \end{align*} We verify the two paths agree. Take $v \in V$ and $\theta \in W^*$. Following the bottom path: \begin{align*} \alpha^{**}(\mathrm{ev}_V(v))(\theta) = \mathrm{ev}_V(v)(\alpha^*(\theta)), \end{align*} by definition of the dual map $\alpha^{**}$ (precompose with $\alpha^*$). Now $\mathrm{ev}_V(v)(\alpha^*(\theta)) = \alpha^*(\theta)(v)$ by definition of the evaluation map, and $\alpha^*(\theta)(v) = \theta(\alpha(v))$ by definition of $\alpha^*$. Following the top path: $\mathrm{ev}_W(\alpha(v))(\theta) = \theta(\alpha(v))$. The two expressions coincide, confirming the diagram commutes. Since $\mathrm{ev}_V$ and $\mathrm{ev}_W$ are isomorphisms (by the [Canonical Isomorphism to Double Dual](/theorems/423)), this says $\alpha^{**} = \mathrm{ev}_W \circ \alpha \circ \mathrm{ev}_V^{-1}$, which under the identification is simply $\alpha^{**} = \alpha$. [/guided] [/step]