[proofplan] The character ring $R(G)$ has a $\mathbb{Z}$-basis given by the irreducible characters $\chi_1, \ldots, \chi_k$ of $G$, so any $\alpha \in R(G)$ admits a unique decomposition $\alpha = \sum_i n_i \chi_i$ with $n_i \in \mathbb{Z}$. The hypothesis $\langle \alpha, \alpha \rangle = 1$ together with orthonormality of the $\chi_i$ in $\mathcal{C}(G)$ ([Row Orthogonality](/theorems/2430)) forces $\sum_i n_i^2 = 1$. The only way to write $1$ as a sum of squares of integers is one term equal to $\pm 1$ and the rest equal to $0$. Hence $\alpha = \pm \chi_{i_0}$ for some $i_0$. Finally, $\alpha(1) > 0$ and $\chi_{i_0}(1) = \dim V_{i_0} > 0$ select the sign $+$, giving $\alpha = \chi_{i_0}$ — an irreducible character. [/proofplan] [step:Decompose $\alpha$ in the basis of irreducible characters] Let $\chi_1, \ldots, \chi_k$ be the distinct irreducible characters of $G$, where $k$ is the number of conjugacy classes of $G$. Recall that the **virtual character ring** (or generalised character ring) \begin{align*} R(G) = \biggl\{\sum_{i=1}^k n_i \chi_i : n_i \in \mathbb{Z}\biggr\} \end{align*} is by definition the $\mathbb{Z}$-span of the irreducible characters inside $\mathcal{C}(G)$, the space of class functions $G \to \mathbb{C}$. Since $\alpha \in R(G)$, there exist integers $n_1, \ldots, n_k \in \mathbb{Z}$ with \begin{align*} \alpha = \sum_{i=1}^{k} n_i \chi_i. \end{align*} The integers $n_i$ are uniquely determined by $\alpha$, since $\chi_1, \ldots, \chi_k$ are linearly independent in $\mathcal{C}(G)$ (being orthonormal, hence linearly independent over $\mathbb{C}$, hence a fortiori over $\mathbb{Z}$). [/step] [step:Use $\langle \alpha, \alpha \rangle = 1$ and orthonormality to get $\sum_i n_i^2 = 1$] By [Row Orthogonality](/theorems/2430), the irreducible characters of $G$ are orthonormal in the $G$-inner product on $\mathcal{C}(G)$: \begin{align*} \langle \chi_i, \chi_j \rangle_G = \delta_{ij}. \end{align*} The inner product on $\mathcal{C}(G)$ is sesquilinear (linear in the first argument, conjugate-linear in the second). Since the $n_i$ are real (in fact integers), $\overline{n_j} = n_j$, and applying linearity in both arguments to $\alpha = \sum_i n_i \chi_i$: \begin{align*} \langle \alpha, \alpha \rangle_G = \biggl\langle \sum_{i=1}^k n_i \chi_i, \sum_{j=1}^k n_j \chi_j \biggr\rangle_G = \sum_{i, j = 1}^{k} n_i\, \overline{n_j}\, \langle \chi_i, \chi_j \rangle_G = \sum_{i, j = 1}^{k} n_i n_j\, \delta_{ij} = \sum_{i=1}^{k} n_i^2. \end{align*} By hypothesis $\langle \alpha, \alpha \rangle_G = 1$, so \begin{align*} \sum_{i=1}^{k} n_i^2 = 1. \end{align*} [/step] [step:Solve $\sum_i n_i^2 = 1$ in integers to identify $\alpha = \pm \chi_{i_0}$] The integers $n_i^2$ are non-negative. The equation $\sum_{i=1}^k n_i^2 = 1$ in non-negative integers admits exactly one solution per index $i_0$: $n_{i_0}^2 = 1$ and $n_i^2 = 0$ for all $i \neq i_0$. (If two or more $n_i$ were nonzero, the sum would exceed $1$ since each squared term is at least $1$; if all were zero, the sum would be $0 \neq 1$.) Hence there exists a unique index $i_0 \in \{1, \ldots, k\}$ with $n_{i_0} \in \{+1, -1\}$, and $n_i = 0$ for $i \neq i_0$. So \begin{align*} \alpha = \pm \chi_{i_0}. \end{align*} [/step] [step:Use $\alpha(1) > 0$ to fix the sign and conclude $\alpha = \chi_{i_0}$] Evaluate at the identity $1 \in G$: \begin{align*} \alpha(1) = \pm \chi_{i_0}(1). \end{align*} The value $\chi_{i_0}(1) = \operatorname{tr}(\rho_{i_0}(1)) = \operatorname{tr}(I_{V_{i_0}}) = \dim V_{i_0}$ is the dimension of the underlying representation space, and dimensions of nonzero vector spaces are strictly positive integers. Hence $\chi_{i_0}(1) > 0$. By hypothesis $\alpha(1) > 0$, so the equation $\alpha(1) = \pm \chi_{i_0}(1)$ with both $\alpha(1), \chi_{i_0}(1) > 0$ forces the $+$ sign: \begin{align*} \alpha = \chi_{i_0}. \end{align*} Thus $\alpha$ is the character of the irreducible representation $\rho_{i_0}: G \to \operatorname{GL}(V_{i_0})$. [guided] The proof is a clean application of three structural facts about the character ring $R(G)$. **First**, $R(G)$ is by construction the free $\mathbb{Z}$-module generated by the irreducible characters $\chi_1, \ldots, \chi_k$. Every $\alpha \in R(G)$ has a unique expression $\alpha = \sum_i n_i \chi_i$ with $n_i \in \mathbb{Z}$. The integers $n_i$ are computed by $n_i = \langle \alpha, \chi_i \rangle_G$ — but we will not need this here. **Second**, the $G$-inner product satisfies orthonormality on irreducibles: $\langle \chi_i, \chi_j \rangle_G = \delta_{ij}$ ([Row Orthogonality](/theorems/2430)). The inner product is conjugate-linear in the second argument, but since the coefficients $n_j$ are integers (hence real), the bilinear expansion has $\overline{n_j} = n_j$: \begin{align*} \langle \alpha, \alpha \rangle_G = \sum_{i, j} n_i n_j \delta_{ij} = \sum_i n_i^2. \end{align*} This identity converts the analytic hypothesis $\langle \alpha, \alpha \rangle_G = 1$ into the arithmetic equation $\sum_i n_i^2 = 1$. **Third**, the equation $\sum_i n_i^2 = 1$ in **integers** has very few solutions: exactly one $n_i$ is $\pm 1$ and the rest are zero. (Compare: in real numbers, infinitely many solutions; the integrality is the key constraint that gives us $\alpha = \pm \chi_{i_0}$ rather than something with smaller coefficients.) This is where the hypothesis $\alpha \in R(G)$ — as opposed to the larger space $\mathcal{C}(G)$ of all class functions — is consumed. The sign is fixed by the second hypothesis. We have $\alpha = \pm \chi_{i_0}$ for some $i_0$, so \begin{align*} \alpha(1) = \pm \chi_{i_0}(1) = \pm \dim V_{i_0}. \end{align*} The dimension of a nonzero representation space is a strictly positive integer, and by hypothesis $\alpha(1) > 0$. Both sides being positive picks out the $+$ sign: $\alpha = \chi_{i_0}$, the character of an irreducible representation. **Why this matters.** This is the standard tool for verifying that an explicitly constructed virtual character (from induction, restriction, tensoring, etc.) is in fact a genuine irreducible character. Computing $\langle \alpha, \alpha \rangle$ and checking $\alpha(1) > 0$ is much more accessible than diagonalising or producing an explicit matrix realisation. [/guided] [/step]