[proofplan] We prove the equivalence of the three conditions by showing (1) $\Rightarrow$ (2) $\Rightarrow$ (1), with (3) being the matrix reformulation of (2). The forward direction uses the adjoint identity to derive $\alpha^*\alpha = \mathrm{id}$, then finite-dimensionality to conclude $\alpha^* = \alpha^{-1}$. The backward direction substitutes $\alpha^*\alpha = \mathrm{id}$ into the adjoint identity. [/proofplan] [step:Show inner-product preservation implies $\alpha^* = \alpha^{-1}$] Suppose $(\alpha(v), \alpha(w)) = (v, w)$ for all $v, w \in V$. By definition of the adjoint, $(\alpha(v), \alpha(w)) = (v, \alpha^*\alpha(w))$. So $(v, w) = (v, \alpha^*\alpha(w))$ for all $v$, giving $(v, w - \alpha^*\alpha(w)) = 0$ for all $v$. By non-degeneracy of the inner product, $\alpha^*\alpha(w) = w$ for all $w$, i.e., $\alpha^*\alpha = \mathrm{id}$. In particular, $\alpha$ is injective: if $\alpha(v) = \mathbf{0}$, then $v = \alpha^*\alpha(v) = \alpha^*(\mathbf{0}) = \mathbf{0}$. Since $V$ is finite-dimensional, an injective endomorphism is bijective, so $\alpha$ is invertible and $\alpha^* = \alpha^{-1}$. [guided] The adjoint identity $(\alpha(v), \alpha(w)) = (v, \alpha^*\alpha(w))$ converts the inner-product-preservation condition into an algebraic equation involving $\alpha^*\alpha$. Setting the two expressions for $(\alpha(v), \alpha(w))$ equal: \begin{align*} (v, w) = (v, \alpha^*\alpha(w)) \quad \text{for all } v, w \in V. \end{align*} This gives $(v, w - \alpha^*\alpha(w)) = 0$ for all $v$. If $w - \alpha^*\alpha(w) \neq \mathbf{0}$, taking $v = w - \alpha^*\alpha(w)$ gives $\|w - \alpha^*\alpha(w)\|^2 = 0$, a contradiction. So $\alpha^*\alpha = \mathrm{id}$. To conclude $\alpha^* = \alpha^{-1}$ (rather than merely a left inverse), we use finite-dimensionality. The equation $\alpha^*\alpha = \mathrm{id}$ shows $\alpha$ is injective. An injective linear map $V \to V$ on a finite-dimensional space is surjective (by the Rank-Nullity theorem), hence invertible. Multiplying $\alpha^*\alpha = \mathrm{id}$ on the right by $\alpha^{-1}$: $\alpha^* = \alpha^{-1}$. [/guided] [/step] [step:Show $\alpha^* = \alpha^{-1}$ implies inner-product preservation] If $\alpha^* = \alpha^{-1}$, then $\alpha^*\alpha = \mathrm{id}$. For any $v, w \in V$: \begin{align*} (\alpha(v), \alpha(w)) = (v, \alpha^*\alpha(w)) = (v, w). \end{align*} [/step] [step:Translate to the matrix characterisation in orthonormal bases] In an orthonormal basis, $\alpha$ has matrix $A$ and $\alpha^*$ has matrix $A^\dagger$. The condition $\alpha^* = \alpha^{-1}$ is equivalent to $A^\dagger A = I$. Over $\mathbb{R}$, $A^\dagger = A^\top$, so this becomes $A^\top A = I$ (orthogonality). Over $\mathbb{C}$, this is $\bar{A}^\top A = I$ (unitarity). [/step]