[proofplan] We prove the identity by comparing Schur coefficients. Since the Schur functions form an [orthonormal basis](/page/Orthonormal%20Basis) for the Hall [inner product](/page/Inner%20Product), it is enough to pair both sides with an arbitrary Schur function $s_\mu$. The adjointness defining $s_\lambda^\perp$ converts this coefficient into the coefficient of $s_\nu$ in the product $s_\lambda s_\mu$, which is the Littlewood-Richardson coefficient $c_{\lambda\mu}^{\nu}$. The Schur expansion of the skew Schur function has exactly these coefficients, so the two symmetric functions agree. [/proofplan] [step:Compute the Schur coefficient of the adjoint expression] Let $\mu$ be an arbitrary partition. Since $s_\lambda^\perp$ is the adjoint of multiplication by $s_\lambda$, we have \begin{align*} (s_\mu, s_\lambda^\perp(s_\nu))_\Lambda &= (s_\lambda s_\mu, s_\nu)_\Lambda. \end{align*} By the [Littlewood-Richardson rule](/theorems/5183), the product $s_\lambda s_\mu$ has Schur expansion \begin{align*} s_\lambda s_\mu &= \sum_{\rho} c_{\lambda\mu}^{\rho} s_\rho, \end{align*} where the sum runs over all partitions $\rho$ and $c_{\lambda\mu}^{\rho}$ is the Littlewood-Richardson coefficient. Pairing with $s_\nu$ and using Schur orthonormality gives \begin{align*} (s_\lambda s_\mu, s_\nu)_\Lambda &= \sum_{\rho} c_{\lambda\mu}^{\rho} (s_\rho, s_\nu)_\Lambda \\ &= c_{\lambda\mu}^{\nu}. \end{align*} Therefore \begin{align*} (s_\mu, s_\lambda^\perp(s_\nu))_\Lambda &= c_{\lambda\mu}^{\nu}. \end{align*} [guided] We want to identify the symmetric function $s_\lambda^\perp(s_\nu)$. The Schur functions form an orthonormal basis of $\Lambda$, so the coefficient of $s_\mu$ in any symmetric function $F \in \Lambda$ is exactly the Hall inner product $(s_\mu,F)_\Lambda$. Thus we fix an arbitrary partition $\mu$ and compute the coefficient of $s_\mu$ in $s_\lambda^\perp(s_\nu)$. By definition, $s_\lambda^\perp$ is the adjoint of the multiplication map \begin{align*} M_{s_\lambda}: \Lambda &\to \Lambda \\ f &\mapsto s_\lambda f. \end{align*} Adjointness means that for all $f,g \in \Lambda$, \begin{align*} (f, s_\lambda^\perp(g))_\Lambda &= (s_\lambda f, g)_\Lambda. \end{align*} Applying this with $f=s_\mu$ and $g=s_\nu$ gives \begin{align*} (s_\mu, s_\lambda^\perp(s_\nu))_\Lambda &= (s_\lambda s_\mu, s_\nu)_\Lambda. \end{align*} Now we expand the product $s_\lambda s_\mu$ in the Schur basis. The Littlewood-Richardson coefficients $c_{\lambda\mu}^{\rho}$ are defined by \begin{align*} s_\lambda s_\mu &= \sum_{\rho} c_{\lambda\mu}^{\rho} s_\rho, \end{align*} where $\rho$ ranges over all partitions. Since the Schur basis is orthonormal for $(\cdot,\cdot)_\Lambda$, we have \begin{align*} (s_\rho, s_\nu)_\Lambda &= \begin{cases} 1, & \rho=\nu,\\ 0, & \rho\neq \nu. \end{cases} \end{align*} Therefore \begin{align*} (s_\lambda s_\mu, s_\nu)_\Lambda &= \left(\sum_{\rho} c_{\lambda\mu}^{\rho} s_\rho, s_\nu\right)_\Lambda \\ &= \sum_{\rho} c_{\lambda\mu}^{\rho}(s_\rho,s_\nu)_\Lambda \\ &= c_{\lambda\mu}^{\nu}. \end{align*} Thus the coefficient of $s_\mu$ in $s_\lambda^\perp(s_\nu)$ is $c_{\lambda\mu}^{\nu}$. [/guided] [/step] [step:Compare with the Schur expansion of the skew Schur function] By the defining Schur expansion of skew Schur functions, \begin{align*} s_{\nu/\lambda} &= \sum_{\mu} c_{\lambda\mu}^{\nu} s_\mu, \end{align*} where the sum runs over all partitions $\mu$. Hence, for every partition $\mu$, \begin{align*} (s_\mu, s_{\nu/\lambda})_\Lambda &= c_{\lambda\mu}^{\nu}. \end{align*} The previous step shows that $s_\lambda^\perp(s_\nu)$ has the same Schur coefficient: \begin{align*} (s_\mu, s_\lambda^\perp(s_\nu))_\Lambda &= (s_\mu, s_{\nu/\lambda})_\Lambda. \end{align*} Since this equality holds for every partition $\mu$ and the Schur functions form a basis of $\Lambda$, the two symmetric functions are equal: \begin{align*} s_\lambda^\perp(s_\nu) &= s_{\nu/\lambda}. \end{align*} [/step] [step:Apply the zero convention when $\lambda$ is not contained in $\nu$] If the Young diagram of $\lambda$ is not contained in the Young diagram of $\nu$, then the skew diagram $\nu/\lambda$ is not a Young-diagram difference. By the stated convention for skew Schur functions in this case, \begin{align*} s_{\nu/\lambda} &=0. \end{align*} The identity already proved gives \begin{align*} s_\lambda^\perp(s_\nu) &=s_{\nu/\lambda} =0. \end{align*} This completes the proof. [/step]