Jacobi–Trudi and dual Jacobi–Trudi identities

For each $n\ge 1$, take the coefficient of $t^n$ in the finite identity [F7] at every rank $N$. By [F4] and [F5], these coefficients are the rank projections of the stable products ${\sum }_{i=0}^n(-1{)}^i{e}_i{h}_{n-i}$. Since all projections vanish, the inverse-limit element is zero by [F1]; the constant coefficient is $e_0{h}_0=1$. Hence $E(-t)H(t)=1$ coefficientwise in $\mathrm{\Lambda }\mathrm{⟦}t\mathrm{⟧}$.

Fix a finite rank $N\ge 1$ and let $e_q^{(k)}$ be the elementary polynomial in the variables other than $x_k$. By [F5] and [F7], $H_N(t){\sum }_{q=0}^{N-1}(-1{)}^q{e}_q^{(k)}{t}^q=(1-x_{k}t{)}^{-1}$. For $\alpha \in {\mathbb{N}}^N$, define $A_{\alpha }=(x_k^{{\alpha }_i}{)}_{i,k}$, $M_{j,k}=(-1{)}^{N-j}{e}_{N-j}^{(k)}$, and $(H_{\alpha }{)}_{i,j}=h_{{\alpha }_i-N+j}$. Taking the coefficient of $t^{{\alpha }_i}$ gives $A_{\alpha }=H_{\alpha }M$.

If $\lambda =\mathrm{\varnothing }$, then $s_{\lambda }=1$ by [F3], and either determinant is upper unitriangular or empty, hence equals $1$. Otherwise fix any finite rank $N\ge r\ge \mathrm{\ell }(\lambda )\ge 1$. Put ${\delta }_N=(N-1,\dots ,0)$ and ${\alpha }_i={\lambda }_i+N-i$. For $\alpha ={\delta }_N$, $H_{{\delta }_N}$ is upper triangular with diagonal $1$, so $\det M=\det A_{{\delta }_N}=a_{{\delta }_N}\mathrm{\ne }0$. For $\alpha =\lambda +{\delta }_N$, $\det A_{\alpha }$ is the bialternant numerator and $H_{\alpha }=(h_{{\lambda }_i-i+j}{)}_{i,j}$. Taking determinants in $A_{\alpha }=H_{\alpha }M$ and using $\det M=\det A_{{\delta }_N}$ gives $\det A_{\alpha }\mathrm{/}\det A_{{\delta }_N}=\det H_{\alpha }$; by [F3] this is the rank-$N$ Schur polynomial. Appending a zero part to $\lambda$ changes the determinant to $\left(\begin{array}{cc}B& v\\ 0& 1\end{array}\right)$, so its value is independent of determinant size; hence for every rank $N\ge r$ the size-$r$ determinant equals the rank-$N$ Schur polynomial. Compatibility gives equality in $\mathrm{\Lambda }$.

For the chosen sizes $r,c$, index matrices by $0,\dots ,r+c-1$ and set $U_{a,b}=h_{b-a}$ and $V_{a,b}=(-1{)}^{b-a}{e}_{b-a}$ when $b\ge a$, with both entries zero when $b<a$. Step 1.1 gives $UV=I$ coefficientwise; both matrices are upper unitriangular, so $V=U^{-1}$ and $\det U=1$.

If $\lambda =\mathrm{\varnothing }$, each determinant is upper unitriangular, or empty, and equals $1$. Otherwise pad $\lambda$ and ${\lambda }^{\mathrm{\prime }}$ with zero parts to the chosen sizes and set $I=\{{\lambda }_i+r-i:1\le i\le r\}$ and $J=\{r-i:1\le i\le r\}$. In increasing order, the minor $U_{J,I}$ is the transpose of $(h_{{\lambda }_i-i+j})$ with both orders reversed, so $\det U_{J,I}=\det (h_{{\lambda }_i-i+j})$. Its complements are $J^c=\{r+j-1:1\le j\le c\}$ and $I^c=\{r-1+j-{\lambda }_j^{\mathrm{\prime }}:1\le j\le c\}$. The listed $I^c$ indices are strictly increasing and lie in $\{0,\dots ,r+c-1\}$. None equals ${\lambda }_i+r-i$, since equality would give ${\lambda }_i+{\lambda }_j^{\mathrm{\prime }}=i+j-1$: if $j\le {\lambda }_i$ the left side is at least $i+j$, and if $j>{\lambda }_i$ it is at most $i+j-2$. The two sets have $r+c$ distinct indices in total and are therefore complementary. At rank $N\ge \mathrm{\ell }(\lambda )$, the bialternant numerator and denominator are nonzero: the strictly decreasing exponents give distinct monomials in the numerator determinant, and the Vandermonde denominator is nonzero. Thus $s_{\lambda }\mathrm{\ne }0$; by [F4], $\mathrm{\Lambda }$ is a domain. Over $K=\mathrm{Frac}(\mathrm{\Lambda })$, $A=U_{J,I}$ is invertible by step 2.1. Reorder rows as $J,J^c$ and columns as $I,I^c$, giving $\tilde{U}=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Block elimination gives $\det \tilde{U}=\det A\det (D-C{A}^{-1}B)$, while the lower-right block of ${\tilde{U}}^{-1}$ is $(D-C{A}^{-1}B{)}^{-1}$. Thus $\det A=\det \tilde{U}\det (V_{I^c,J^c})$. Moving an increasing index set $S$ of size $r$ to the front has sign $(-1{)}^{\sum S-r(r-1)\mathrm{/}2}$; the row and column reorderings therefore give $\det \tilde{U}=(-1{)}^{\sum I+\sum J}\det U$. Since $\sum I+\sum J=\mathrm{\mid }\lambda \mathrm{\mid }+r(r-1)$ and $\det U=1$, this sign is $(-1{)}^{\mathrm{\mid }\lambda \mathrm{\mid }}$. The complementary minor has entries $V_{r-1+i-{\lambda }_i^{\mathrm{\prime }},r+j-1}=(-1{)}^{{\lambda }_i^{\mathrm{\prime }}-i+j}{e}_{{\lambda }_i^{\mathrm{\prime }}-i+j}$; a negative subscript gives zero by the stated convention. Factoring row and column signs contributes $(-1{)}^{{\sum }_i{\lambda }_i^{\mathrm{\prime }}-{\sum }_{i}i+{\sum }_{j}j}=(-1{)}^{\mathrm{\mid }\lambda \mathrm{\mid }}$, which cancels the permutation sign. Therefore $\det (h_{{\lambda }_i-i+j})=\det (e_{{\lambda }_i^{\mathrm{\prime }}-i+j})$. For $\lambda =(1)$ and $r=c=1$, these are $h_1$ and $e_1$, each the sum of the variables by [F5] and [F6].