[proofplan] Part (1) is an element chase: $\ker\alpha^* = (\mathrm{im}\,\alpha)^0$. Part (3) follows from part (1) by dimension counting using [Rank-Nullity](/theorems/384) and [Dimension of Annihilator](/theorems/420). Part (2) uses containment $\mathrm{im}\,\alpha^* \subseteq (\ker\alpha)^0$ plus a dimension count showing equality. [/proofplan] [step:Show $\ker\alpha^* = (\mathrm{im}\,\alpha)^0$ by unwinding definitions] A functional $\theta \in W^*$ lies in $\ker\alpha^*$ iff $\alpha^*(\theta) = 0$ iff $\theta \circ \alpha = 0$ iff $\theta(\alpha(v)) = 0$ for all $v \in V$. This is exactly the condition that $\theta$ vanishes on $\mathrm{im}\,\alpha$, i.e., $\theta \in (\mathrm{im}\,\alpha)^0$. [/step] [step:Prove $\mathrm{rank}(\alpha) = \mathrm{rank}(\alpha^*)$ by dimension counting] By part (1) and the [Dimension of Annihilator](/theorems/420) applied to $\mathrm{im}\,\alpha \subseteq W$: \begin{align*} \dim\ker\alpha^* = \dim(\mathrm{im}\,\alpha)^0 = \dim W - \dim(\mathrm{im}\,\alpha) = \dim W - \mathrm{rank}(\alpha). \end{align*} By [Rank-Nullity](/theorems/384) applied to $\alpha^*: W^* \to V^*$: \begin{align*} \mathrm{rank}(\alpha^*) = \dim W^* - \dim\ker\alpha^*. \end{align*} Since $\dim W^* = \dim W$ by the [Dimension of Dual Space](/theorems/415): \begin{align*} \mathrm{rank}(\alpha^*) = \dim W - (\dim W - \mathrm{rank}(\alpha)) = \mathrm{rank}(\alpha). \end{align*} [/step] [step:Show $\mathrm{im}\,\alpha^* = (\ker\alpha)^0$ by containment and dimension] **Containment $\mathrm{im}\,\alpha^* \subseteq (\ker\alpha)^0$:** If $\phi = \alpha^*(\theta) = \theta \circ \alpha$ and $v \in \ker\alpha$, then $\phi(v) = \theta(\alpha(v)) = \theta(\mathbf{0}) = 0$. So $\mathrm{im}\,\alpha^* \subseteq (\ker\alpha)^0$. **Equality by dimension:** $\dim\mathrm{im}\,\alpha^* = \mathrm{rank}(\alpha^*) = \mathrm{rank}(\alpha)$ by the previous step. By [Dimension of Annihilator](/theorems/420) and [Rank-Nullity](/theorems/384): \begin{align*} \dim(\ker\alpha)^0 = \dim V - \dim\ker\alpha = \mathrm{rank}(\alpha). \end{align*} Since $\mathrm{im}\,\alpha^* \subseteq (\ker\alpha)^0$ and both have dimension $\mathrm{rank}(\alpha)$, they are equal. [/step]