[proofplan] The proof has two directions, both routed through the orthogonality of irreducible characters. For the forward direction, irreducibility of $\rho$ implies $\chi$ is itself an irreducible character, so [Row Orthogonality](/theorems/2430) gives $\langle \chi, \chi \rangle = 1$ directly. For the converse, [Maschke's Theorem](/theorems/2409) decomposes $\rho$ into irreducibles, $\chi = \sum_j m_j \chi_j$ with $m_j \in \mathbb{Z}_{\geq 0}$, and orthonormality of the $\chi_j$ gives $\langle \chi, \chi \rangle = \sum_j m_j^2$. The Diophantine equation $\sum_j m_j^2 = 1$ in non-negative integers forces exactly one $m_j$ to equal $1$ and the rest to vanish, so $\chi$ is irreducible. [/proofplan] [step:Forward direction: irreducibility of $\rho$ gives $\langle \chi, \chi \rangle = 1$] Suppose $\rho: G \to \operatorname{GL}(V)$ is an irreducible complex representation with character $\chi$. Then $\chi$ is by definition the character of an irreducible representation, so it is one of the irreducible characters $\chi_1, \ldots, \chi_k$ of $G$. Apply the [Row Orthogonality Relations](/theorems/2430) with $\chi' = \chi$: since $\rho$ is isomorphic to itself, \begin{align*} \langle \chi, \chi \rangle = 1. \end{align*} [/step] [step:Converse: assume $\langle \chi, \chi \rangle = 1$ and decompose $\rho$ into irreducibles via Maschke] For the converse, suppose $\chi$ is the character of a complex representation $\rho: G \to \operatorname{GL}(V)$ and that $\langle \chi, \chi \rangle = 1$. Apply [Maschke's Theorem](/theorems/2409) to $\rho$. The hypotheses are: $G$ is finite (global hypothesis of the theorem) and $\operatorname{char} \mathbb{C} = 0$, so $\operatorname{char} \mathbb{C} \nmid |G|$. Hence $\rho$ is completely reducible. Let $\rho_1, \ldots, \rho_k$ be a complete list of representatives of the isomorphism classes of irreducible complex representations of $G$, with characters $\chi_1, \ldots, \chi_k$. There exist non-negative integers $m_1, \ldots, m_k \in \mathbb{Z}_{\geq 0}$ and a $G$-equivariant isomorphism \begin{align*} V \cong \bigoplus_{j=1}^k V_j^{\oplus m_j}, \end{align*} where $V_j$ is the underlying space of $\rho_j$. Equivalently, $\rho \cong \bigoplus_j m_j \rho_j$. By Property 4 of [Elementary Properties of Characters](/theorems/2421) (additivity of characters under direct sums), iterated: \begin{align*} \chi = \sum_{j=1}^k m_j \chi_j. \end{align*} [guided] The strategy is to convert the assumption $\langle \chi, \chi \rangle = 1$ into a statement about the multiplicities of irreducible constituents of $\rho$. The first step is to write down the multiplicities at all. We invoke [Maschke's Theorem](/theorems/2409). It requires the field characteristic not to divide $|G|$. Here the field is $\mathbb{C}$ with characteristic $0$, and $0 \nmid n$ for any positive integer $n$, so the hypothesis is satisfied. Maschke produces a decomposition of $V$ into a direct sum of $G$-invariant irreducible subspaces, which we group by isomorphism class: \begin{align*} V \cong \bigoplus_{j=1}^k V_j^{\oplus m_j}. \end{align*} The integer $m_j \geq 0$ is the **multiplicity** of $\rho_j$ in $\rho$. The character of a direct sum is the sum of characters (Property 4 of [Elementary Properties of Characters](/theorems/2421)). Applying this to the decomposition above gives \begin{align*} \chi = \sum_{j=1}^k m_j \chi_j. \end{align*} Now the inner product $\langle \chi, \chi \rangle$ becomes a sum we can evaluate using orthogonality of the $\chi_j$. [/guided] [/step] [step:Expand $\langle \chi, \chi \rangle$ using row orthogonality] By the [Row Orthogonality Relations](/theorems/2430), the irreducible characters satisfy \begin{align*} \langle \chi_i, \chi_j \rangle = \delta_{ij} \quad \text{for } 1 \leq i, j \leq k. \end{align*} Expand $\langle \chi, \chi \rangle$ using sesquilinearity of the inner product, with $\chi = \sum_j m_j \chi_j$ on both sides: \begin{align*} \langle \chi, \chi \rangle = \left\langle \sum_{i=1}^k m_i \chi_i,\; \sum_{j=1}^k m_j \chi_j \right\rangle = \sum_{i,j=1}^k m_i \overline{m_j}\, \langle \chi_i, \chi_j \rangle = \sum_{i,j=1}^k m_i m_j\, \delta_{ij} = \sum_{j=1}^k m_j^2. \end{align*} We used $\overline{m_j} = m_j$ because the multiplicities $m_j$ are integers, hence real. [guided] We have $\chi = \sum_j m_j \chi_j$ and we want to compute $\langle \chi, \chi \rangle$. The inner product is sesquilinear (linear in the first slot, conjugate-linear in the second) but the multiplicities $m_j$ are non-negative integers, hence real, so conjugation does nothing to them. Expanding the bilinear sum: \begin{align*} \langle \chi, \chi \rangle = \sum_{i,j} m_i m_j \langle \chi_i, \chi_j \rangle. \end{align*} Row orthogonality ([Row Orthogonality Relations](/theorems/2430)) says $\langle \chi_i, \chi_j \rangle = \delta_{ij}$: irreducible characters are an orthonormal family in the space $\mathcal{C}(G)$ of class functions. The off-diagonal terms vanish, and the diagonal terms collapse to \begin{align*} \langle \chi, \chi \rangle = \sum_j m_j^2. \end{align*} This is the master identity: the squared norm of $\chi$ equals the sum of squared multiplicities of its irreducible constituents. [/guided] [/step] [step:Solve the Diophantine equation $\sum_j m_j^2 = 1$ in non-negative integers] Combine the hypothesis $\langle \chi, \chi \rangle = 1$ with the formula from the previous step: \begin{align*} \sum_{j=1}^k m_j^2 = 1, \quad m_j \in \mathbb{Z}_{\geq 0}. \end{align*} Each $m_j^2$ is a non-negative integer. The only way a sum of non-negative integers equals $1$ is if exactly one summand is $1$ and all others vanish. Since $m_j^2 = 1$ with $m_j \geq 0$ forces $m_j = 1$, we conclude: there is a unique index $j_0 \in \{1, \ldots, k\}$ with $m_{j_0} = 1$ and $m_j = 0$ for all $j \neq j_0$. Substituting back: \begin{align*} \chi = m_{j_0} \chi_{j_0} = \chi_{j_0}, \end{align*} and the decomposition reads $\rho \cong \rho_{j_0}$. Hence $\rho$ is irreducible. [guided] We have reached \begin{align*} 1 = \langle \chi, \chi \rangle = \sum_j m_j^2. \end{align*} The right-hand side is a sum of squares of non-negative integers. We are looking for non-negative integer solutions of $\sum_j m_j^2 = 1$. Each term $m_j^2$ is in $\{0, 1, 4, 9, 16, \ldots\}$. For the sum to be $1$, exactly one term must equal $1$ and all others must equal $0$. The unique non-negative integer with $m_j^2 = 1$ is $m_j = 1$. So there is exactly one index $j_0$ with $m_{j_0} = 1$, and $m_j = 0$ for $j \neq j_0$. The decomposition $\rho \cong \bigoplus_j m_j \rho_j$ collapses to $\rho \cong \rho_{j_0}$, a single irreducible. The character correspondingly reduces to $\chi = \chi_{j_0}$, an irreducible character. The argument is purely arithmetic at the end: irreducibility is detected by an integrality/positivity constraint on a single rational number, $\langle \chi, \chi \rangle$. This is the value of the criterion — irreducibility, normally a structural statement about subrepresentations, becomes a single number to compute. [/guided] [/step] [step:Combine both directions] Step 1 establishes that $\rho$ irreducible implies $\langle \chi, \chi \rangle = 1$. Steps 2 to 4 establish the converse: $\langle \chi, \chi \rangle = 1$ implies $\rho$ is isomorphic to a single irreducible $\rho_{j_0}$, hence irreducible. Together these give the biconditional in the statement. [/step]