[proofplan] The orthogonality formula is the defining normalization of the Hall [inner product](/page/Inner%20Product) on the power sum basis. To prove nondegeneracy in homogeneous degree $n$, we use that the functions $p_\lambda$ with $\lambda \vdash n$ form a basis of $\operatorname{Sym}^n$. If a homogeneous element pairs to zero with every element of $\operatorname{Sym}^n$, then pairing it with each basis vector $p_\mu$ isolates one coefficient because the Gram matrix is diagonal with positive diagonal entries $z_\mu$. [/proofplan] [step:Apply the defining rule of the Hall inner product to the power sum basis] By definition of the Hall inner product on $\operatorname{Sym}$, the power sum symmetric functions satisfy \begin{align*} \langle p_\lambda, p_\mu \rangle = z_\lambda \delta_{\lambda,\mu} \end{align*} for all partitions $\lambda$ and $\mu$, where \begin{align*} z_\lambda = \prod_{i \geq 1} i^{m_i(\lambda)} m_i(\lambda)!. \end{align*} This proves the asserted orthogonality formula. [/step] [step:Expand a homogeneous element in the power sum basis] Fix $n \in \mathbb{N}$. Let $f \in \operatorname{Sym}^n$ be an element such that \begin{align*} \langle f,g\rangle = 0 \end{align*} for every $g \in \operatorname{Sym}^n$. Since $\{p_\lambda : \lambda \vdash n\}$ is a basis of $\operatorname{Sym}^n$, there are unique coefficients $a_\lambda \in \mathbb{Q}$, indexed by partitions $\lambda$ of $n$, such that \begin{align*} f = \sum_{\lambda \vdash n} a_\lambda p_\lambda. \end{align*} [guided] Fix $n \in \mathbb{N}$. To prove nondegeneracy on $\operatorname{Sym}^n$, we must show that the only element of $\operatorname{Sym}^n$ orthogonal to every element of $\operatorname{Sym}^n$ is the zero element. Thus let $f \in \operatorname{Sym}^n$ satisfy \begin{align*} \langle f,g\rangle = 0 \end{align*} for every $g \in \operatorname{Sym}^n$. The useful basis in this argument is the power sum basis. Since $\{p_\lambda : \lambda \vdash n\}$ is a basis of the homogeneous component $\operatorname{Sym}^n$, the element $f$ has a unique expansion \begin{align*} f = \sum_{\lambda \vdash n} a_\lambda p_\lambda, \end{align*} where each coefficient $a_\lambda$ belongs to $\mathbb{Q}$. This expansion is the point at which orthogonality becomes useful: pairing $f$ with a single $p_\mu$ will select exactly the coefficient indexed by $\mu$. [/guided] [/step] [step:Isolate each coefficient by pairing with one power sum] Let $\mu$ be a partition of $n$. Since $p_\mu \in \operatorname{Sym}^n$, the hypothesis gives \begin{align*} 0 = \langle f,p_\mu\rangle. \end{align*} Using linearity of the Hall inner product in the first argument and the orthogonality formula already proved, \begin{align*} 0 = \langle f,p_\mu\rangle = \left\langle \sum_{\lambda \vdash n} a_\lambda p_\lambda,p_\mu \right\rangle = \sum_{\lambda \vdash n} a_\lambda \langle p_\lambda,p_\mu\rangle = \sum_{\lambda \vdash n} a_\lambda z_\lambda \delta_{\lambda,\mu} = a_\mu z_\mu. \end{align*} For every partition $\mu$, the integer \begin{align*} z_\mu = \prod_{i \geq 1} i^{m_i(\mu)} m_i(\mu)! \end{align*} is positive. Hence $a_\mu z_\mu = 0$ implies $a_\mu = 0$. [/step] [step:Conclude nondegeneracy on each homogeneous component] Since the partition $\mu \vdash n$ was arbitrary, all coefficients $a_\mu$ in the power sum expansion of $f$ vanish. Therefore \begin{align*} f = \sum_{\mu \vdash n} 0 \cdot p_\mu = 0. \end{align*} Thus no nonzero element of $\operatorname{Sym}^n$ is orthogonal to all of $\operatorname{Sym}^n$, so the restriction of the Hall inner product to $\operatorname{Sym}^n$ is nondegenerate. [/step]