[proofplan] Over $\mathbb{C}$, the characteristic polynomial splits, so eigenvalues exist. We show they lie on the unit circle using inner-product preservation, then show eigenvectors for distinct eigenvalues are orthogonal. The inductive step uses the invariance of $\langle e_1 \rangle^\perp$ under $\alpha$, established by the relation $\alpha^*(e_1) = \bar{\lambda}_1 e_1$. [/proofplan] [step:Show eigenvalues lie on the unit circle using inner-product preservation] Let $\alpha(v) = \lambdav$ with $v \neq \mathbf{0}$. Since $\alpha$ is unitary ($\alpha^* = \alpha^{-1}$), $\alpha$ preserves the inner product: \begin{align*} \|v\|^2 = (\alpha(v), \alpha(v)) = (\lambdav, \lambdav) = |\lambda|^2\,(v, v) = |\lambda|^2\,\|v\|^2. \end{align*} Since $\|v\|^2 > 0$: $|\lambda|^2 = 1$, i.e., $|\lambda| = 1$. [/step] [step:Show eigenvectors for distinct eigenvalues are orthogonal] If $\alpha(v) = \lambdav$ and $\alpha(w) = \muw$ with $\lambda \neq \mu$, inner-product preservation gives: \begin{align*} (v, w) = (\alpha(v), \alpha(w)) = (\lambdav, \muw) = \bar{\lambda}\mu\,(v, w). \end{align*} If $(v, w) \neq 0$, then $\bar{\lambda}\mu = 1$, so $\mu = 1/\bar{\lambda}$. Since $|\lambda| = 1$, $1/\bar{\lambda} = \lambda$, giving $\mu = \lambda$, contradicting $\lambda \neq \mu$. Therefore $(v, w) = 0$. [/step] [step:Prove $\alpha^*(e_1) = \bar{\lambda}_1e_1$ and conclude $\langle e_1 \rangle^\perp$ is invariant] Over $\mathbb{C}$, the characteristic polynomial $\chi_\alpha$ has degree $n \geq 1$ and splits, so $\alpha$ has an eigenvalue $\lambda_1$ with unit eigenvector $e_1$. Let $W = \langle e_1 \rangle^\perp$. Since $\alpha^* = \alpha^{-1}$ and $\alpha(e_1) = \lambda_1e_1$: \begin{align*} \alpha^*(e_1) = \alpha^{-1}(e_1) = \lambda_1^{-1}e_1 = \bar{\lambda}_1e_1, \end{align*} where $\lambda_1^{-1} = \bar{\lambda}_1$ because $|\lambda_1| = 1$. For any $w \in W$: \begin{align*} (e_1, \alpha(w)) = (\alpha^*(e_1), w) = (\bar{\lambda}_1e_1, w) = \lambda_1\,(e_1, w) = \lambda_1 \cdot 0 = 0. \end{align*} So $\alpha(w) \in W$, proving $\alpha(W) \subseteq W$. [guided] The invariance of $W = \langle e_1 \rangle^\perp$ under $\alpha$ is the key step that makes the induction work. The argument requires knowing how $\alpha^*$ acts on the eigenvector $e_1$. Since $\alpha(e_1) = \lambda_1 e_1$ and $\alpha$ is invertible (being unitary), $\alpha^{-1}(e_1) = \lambda_1^{-1}e_1$. The unitarity condition $\alpha^* = \alpha^{-1}$ then gives $\alpha^*(e_1) = \lambda_1^{-1}e_1$. Since $|\lambda_1| = 1$, $\lambda_1^{-1} = \bar{\lambda}_1$. Now check invariance. For $w \in W$, we need $(e_1, \alpha(w)) = 0$. By the adjoint identity: \begin{align*} (e_1, \alpha(w)) = (\alpha^*(e_1), w) = (\bar{\lambda}_1e_1, w) = \lambda_1\,(e_1, w). \end{align*} The last equality pulls $\bar{\lambda}_1$ out of the conjugate-linear first slot, giving $\overline{\bar{\lambda}_1} = \lambda_1$. Since $w \in W = \langle e_1 \rangle^\perp$, $(e_1, w) = 0$, so $(e_1, \alpha(w)) = 0$. [/guided] [/step] [step:Complete the induction to obtain an orthonormal eigenbasis] The restriction $\alpha|_W: W \to W$ is unitary on the $(n-1)$-dimensional inner product space $W$: for $w_1, w_2 \in W$, $(\alpha(w_1), \alpha(w_2)) = (w_1, w_2)$ since inner-product preservation holds on all of $V$. By induction, $W$ has an orthonormal eigenbasis $(e_2, \dots, e_n)$ for $\alpha|_W$. Then $(e_1, e_2, \dots, e_n)$ is an orthonormal eigenbasis for $\alpha$ on $V$. [/step]