[proofplan] We prove the identity by testing both sides against every element of the power-sum basis. The adjoint definition converts the left-hand pairing into the Hall [inner product](/page/Inner%20Product) of $p_\lambda$ with $p_kp_\mu$, and power-sum orthogonality makes this nonzero exactly when $\lambda$ is obtained from $\mu$ by adjoining one part $k$. In that case the scalar is computed from the explicit formula for $z_\lambda$, and nondegeneracy of the Hall inner product on the power-sum basis identifies the two symmetric functions. [/proofplan]

[step:Pair the adjoint against an arbitrary power-sum basis element]Let $\mu=(1^{m_1(\mu)}2^{m_2(\mu)}\cdots)$ be an arbitrary partition. Since $p_k^\perp$ is the adjoint of the multiplication map \begin{align*} M_{p_k}:\Lambda_{\mathbb{Q}} &\to \Lambda_{\mathbb{Q}}\\ f &\mapsto p_k f, \end{align*} we have \begin{align*} \langle p_k^\perp p_\lambda, p_\mu\rangle = \langle p_\lambda, p_k p_\mu\rangle. \end{align*} The product of power sums is indexed by multiset union of parts, so \begin{align*} p_k p_\mu = p_{\mu\cup k}, \end{align*} where $\mu\cup k$ denotes the partition obtained from $\mu$ by adjoining one part equal to $k$. Therefore the defining orthogonality relation for the Hall inner product gives \begin{align*} \langle p_k^\perp p_\lambda, p_\mu\rangle = \langle p_\lambda, p_{\mu\cup k}\rangle = \begin{cases} z_\lambda, & \lambda=\mu\cup k,\\ 0, & \lambda\neq \mu\cup k. \end{cases} \end{align*}[/step]

[guided]We test against $p_\mu$ because the family $\{p_\mu\}$ is the basis in which the Hall inner product is diagonal. Let $\mu=(1^{m_1(\mu)}2^{m_2(\mu)}\cdots)$ be an arbitrary partition. The operator $p_k^\perp$ is defined as the adjoint of multiplication by $p_k$, meaning precisely that for every $f,g\in\Lambda_{\mathbb{Q}}$, \begin{align*} \langle p_k^\perp f,g\rangle=\langle f,p_k g\rangle. \end{align*} Applying this with $f=p_\lambda$ and $g=p_\mu$ gives \begin{align*} \langle p_k^\perp p_\lambda, p_\mu\rangle = \langle p_\lambda, p_k p_\mu\rangle. \end{align*} Now multiplication of power sums simply concatenates parts. If $\mu$ has multiplicities $m_i(\mu)$, then $p_kp_\mu$ has multiplicities $m_k(\mu)+1$ at $k$ and $m_i(\mu)$ at every $i\neq k$. Thus \begin{align*} p_kp_\mu=p_{\mu\cup k}, \end{align*} where $\mu\cup k$ is the partition obtained by adjoining one part $k$. The Hall inner product is diagonal on the power-sum basis: \begin{align*} \langle p_\alpha,p_\beta\rangle=\delta_{\alpha,\beta}z_\alpha. \end{align*} Using this with $\alpha=\lambda$ and $\beta=\mu\cup k$ yields \begin{align*} \langle p_k^\perp p_\lambda, p_\mu\rangle = \langle p_\lambda, p_{\mu\cup k}\rangle = \begin{cases} z_\lambda, & \lambda=\mu\cup k,\\ 0, & \lambda\neq \mu\cup k. \end{cases} \end{align*} This is the whole point of passing to pairings: the adjoint moves $p_k^\perp$ off the unknown expression, and orthogonality reduces the computation to a single possible partition.[/guided]

[step:Compute the pairing of the proposed right-hand side] First suppose $m_k(\lambda)=0$. Then no partition $\mu$ satisfies $\lambda=\mu\cup k$, so the previous step gives \begin{align*} \langle p_k^\perp p_\lambda,p_\mu\rangle=0 \end{align*} for every partition $\mu$, matching the proposed value $0$. Now suppose $m_k(\lambda)>0$, and let $\lambda\setminus k$ be the partition whose multiplicities are \begin{align*} m_i(\lambda\setminus k) = \begin{cases} m_k(\lambda)-1, & i=k,\\ m_i(\lambda), & i\neq k. \end{cases} \end{align*} The proposed right-hand side is \begin{align*} R:=k\,m_k(\lambda)\,p_{\lambda\setminus k}. \end{align*} For every partition $\mu$, \begin{align*} \langle R,p_\mu\rangle = k\,m_k(\lambda)\,\langle p_{\lambda\setminus k},p_\mu\rangle = \begin{cases} k\,m_k(\lambda)\,z_{\lambda\setminus k}, & \mu=\lambda\setminus k,\\ 0, & \mu\neq \lambda\setminus k. \end{cases} \end{align*} Using the definition of $z_\lambda$ and separating the factor indexed by $k$, \begin{align*} z_\lambda &= \prod_{i\geq 1} i^{m_i(\lambda)}m_i(\lambda)!\\ &= k^{m_k(\lambda)}m_k(\lambda)!\prod_{i\neq k}i^{m_i(\lambda)}m_i(\lambda)!\\ &= k\,m_k(\lambda)\, k^{m_k(\lambda)-1}(m_k(\lambda)-1)!\prod_{i\neq k}i^{m_i(\lambda)}m_i(\lambda)!\\ &= k\,m_k(\lambda)\,z_{\lambda\setminus k}. \end{align*} Thus \begin{align*} \langle R,p_\mu\rangle = \begin{cases} z_\lambda, & \mu=\lambda\setminus k,\\ 0, & \mu\neq \lambda\setminus k. \end{cases} \end{align*} Since $\lambda=\mu\cup k$ is equivalent to $\mu=\lambda\setminus k$ when $m_k(\lambda)>0$, this is exactly the same pairing computed for $p_k^\perp p_\lambda$. [/step]

[step:Use nondegeneracy of the Hall inner product to identify the symmetric functions] If $m_k(\lambda)=0$, the first step gives \begin{align*} \langle p_k^\perp p_\lambda,p_\mu\rangle=0 \end{align*} for every partition $\mu$. Since the power sums form a basis of $\Lambda_{\mathbb{Q}}$ and the Hall inner product satisfies $\langle p_\mu,p_\mu\rangle=z_\mu\neq 0$ for every partition $\mu$, the inner product is nondegenerate. Hence $p_k^\perp p_\lambda=0$. If $m_k(\lambda)>0$, the previous step shows that \begin{align*} \langle p_k^\perp p_\lambda,p_\mu\rangle = \langle k\,m_k(\lambda)\,p_{\lambda\setminus k},p_\mu\rangle \end{align*} for every partition $\mu$. By the same nondegeneracy, the two symmetric functions are equal: \begin{align*} p_k^\perp p_\lambda = k\,m_k(\lambda)\,p_{\lambda\setminus k}. \end{align*} Combining the two cases proves the stated formula. [/step]