[proofplan] We prove the tropical Plücker relations directly from the ordinary determinant identities satisfied by maximal minors. First we derive, using Laplace expansion and exterior products, the three-term-or-more Plücker relation attached to an $(r-1)$-set $S$ and an $(r+1)$-set $T$. Then we apply the non-Archimedean valuation inequality in the valued field $K$ to the resulting zero-sum relation: a finite sum equal to zero cannot have a unique term of strictly smallest valuation. Since $A$ has rank $r$, at least one maximal minor is nonzero, so the resulting tropical Plücker vector is not identically $\infty$. [/proofplan]

[step:Define the maximal minors and record that at least one is finite] Let $\Gamma$ denote the totally ordered abelian group in which the nonzero values of the valuation lie. Let $\nu: K \to \Gamma \cup \{\infty\}$ denote the valuation on $K$, extended by the convention $\nu(0)=\infty$. Let $[n] := \{1,\dots,n\}$, let $E := K^r$, and for each $j \in [n]$ let $a_j \in E$ denote the $j$-th column of $A$. Define the maximal-minor map \begin{align*} \Delta: \binom{[n]}{r} \to K, \qquad B \mapsto \det A_B. \end{align*} Then $p(B)=\nu(\Delta(B))$ for every $B \in \binom{[n]}{r}$. Since $\operatorname{rank} A=r$, there exists at least one $r$-element subset $B_0 \subset [n]$ such that $\Delta(B_0)\ne 0$. Hence $p(B_0)\in \Gamma$, so $p$ is not identically $\infty$. [/step]

[step:Derive the Plücker relation for one pair of index sets]Fix an $(r-1)$-element subset $S \subset [n]$ and an $(r+1)$-element subset $T \subset [n]$. Write \begin{align*} T=\{t_1<\cdots<t_{r+1}\}. \end{align*} For each $k \in \{1,\dots,r+1\}$, define \begin{align*} C_k := \Delta(T\setminus\{t_k\}) \in K. \end{align*} We first claim that the following vector identity holds in $E$: \begin{align*} \sum_{k=1}^{r+1}(-1)^{k-1}C_k a_{t_k}=0. \end{align*} Indeed, for a fixed coordinate index $q \in \{1,\dots,r\}$, the $q$-th coordinate of this vector is \begin{align*} \sum_{k=1}^{r+1}(-1)^{k-1}C_k (a_{t_k})_q. \end{align*} This is the [Laplace expansion](/page/Determinant) along the first row of the determinant of the $(r+1)\times(r+1)$ matrix obtained by placing the $q$-th row of the $r\times(r+1)$ matrix $(a_{t_1},\dots,a_{t_{r+1}})$ above that same matrix, after using the same column order $t_1<\cdots<t_{r+1}$ in every maximal minor. The cofactor sign of the entry in column $k$ is $(-1)^{1+k}$, which equals $(-1)^{k-1}$, so the displayed alternating signs are exactly the Laplace signs. That determinant is zero because its first row is equal to the row of the same matrix indexed by $q+1$. Since this holds for every coordinate $q$, the vector identity follows. Let $\omega := e_1\wedge\cdots\wedge e_r \in \bigwedge^r E$, where $e_1,\dots,e_r$ is the standard basis of $E$. Write \begin{align*} S=\{s_1<\cdots<s_{r-1}\}. \end{align*} Wedge the vector identity on the left with $a_{s_1}\wedge\cdots\wedge a_{s_{r-1}}$. If $t_k \in S$, then the corresponding exterior product contains a repeated vector and is therefore zero. If $t_k\in T\setminus S$, then there is a sign $\varepsilon_k\in\{1,-1\}$ such that \begin{align*} a_{s_1}\wedge\cdots\wedge a_{s_{r-1}}\wedge a_{t_k}=\varepsilon_k\Delta(S\cup\{t_k\})\omega. \end{align*} Thus the coefficient of $\omega$ in the wedged identity gives \begin{align*} \sum_{t\in T\setminus S}\sigma_t \Delta(S\cup\{t\})\Delta(T\setminus\{t\})=0 \end{align*} for signs $\sigma_t\in\{1,-1\}$ depending only on the relative order of the indices.[/step]

[guided]We derive the ordinary Pluecker relation in a form whose signs will not matter after applying the valuation. Fix an $(r-1)$-element subset $S\subset[n]$ and an $(r+1)$-element subset $T\subset[n]$, and write \begin{align*} T=\{t_1<\cdots<t_{r+1}\}. \end{align*} For each $k\in\{1,\dots,r+1\}$, define \begin{align*} C_k:=\Delta(T\setminus\{t_k\})\in K. \end{align*} We first prove the column relation \begin{align*} \sum_{k=1}^{r+1}(-1)^{k-1}C_k a_{t_k}=0 \end{align*} in $E=K^r$. To verify a vector identity in $K^r$, it is enough to verify each coordinate. Fix $q\in\{1,\dots,r\}$. The $q$-th coordinate of the left-hand side is \begin{align*} \sum_{k=1}^{r+1}(-1)^{k-1}C_k(a_{t_k})_q. \end{align*} Consider the $(r+1)\times(r+1)$ matrix whose first row is the $q$-th row of $(a_{t_1},\dots,a_{t_{r+1}})$ and whose remaining $r$ rows are the rows of $(a_{t_1},\dots,a_{t_{r+1}})$ in their original order. Its determinant is zero because the first row equals the row indexed by $q+1$. Expanding this determinant along the first row by [Laplace expansion](/page/Determinant), the cofactor sign in column $k$ is $(-1)^{1+k}=(-1)^{k-1}$, and the corresponding minor is exactly $\Delta(T\setminus\{t_k\})=C_k$ with the chosen column order. Therefore the displayed coordinate sum is zero. Since this holds for all $q$, the column relation follows. Now write \begin{align*} S=\{s_1<\cdots<s_{r-1}\}. \end{align*} Let $\omega:=e_1\wedge\cdots\wedge e_r\in\bigwedge^rE$, where $e_1,\dots,e_r$ is the standard basis of $E$. Wedge the column relation on the left with $a_{s_1}\wedge\cdots\wedge a_{s_{r-1}}$. If $t_k\in S$, then the exterior product contains the repeated vector $a_{t_k}$, hence is zero by alternation of the exterior product. If $t_k\in T\setminus S$, then $S\cup\{t_k\}$ is an $r$-element subset of $[n]$, and reordering the wedge product into increasing order introduces a sign $\varepsilon_k\in\{1,-1\}$ satisfying \begin{align*} a_{s_1}\wedge\cdots\wedge a_{s_{r-1}}\wedge a_{t_k}=\varepsilon_k\Delta(S\cup\{t_k\})\omega. \end{align*} Taking the coefficient of $\omega$ in the wedged identity gives \begin{align*} \sum_{t\in T\setminus S}\sigma_t\Delta(S\cup\{t\})\Delta(T\setminus\{t\})=0, \end{align*} where each $\sigma_t\in\{1,-1\}$ is the product of the Laplace sign and the reordering sign. This is the ordinary Pluecker relation needed for the valuation argument.[/guided]

[step:Convert the Plücker relation into the tropical minimum condition]For each $t\in T\setminus S$, define \begin{align*} c_t:=\sigma_t \Delta(S\cup\{t\})\Delta(T\setminus\{t\}) \in K. \end{align*} The previous step gives \begin{align*} \sum_{t\in T\setminus S}c_t=0. \end{align*} Since $\sigma_t\in\{1,-1\}$ is nonzero, $\nu(\sigma_t)=0$. Therefore \begin{align*} \nu(c_t)=p(S\cup\{t\})+p(T\setminus\{t\}) \end{align*} for every $t\in T\setminus S$, with the convention that $\infty+\gamma=\infty$. We show that the minimum of these values cannot be attained at a unique index. Suppose, for contradiction, that there is a unique $t_0\in T\setminus S$ such that \begin{align*} \nu(c_{t_0})<\nu(c_t) \end{align*} for every $t\in T\setminus S$ with $t\ne t_0$. Then \begin{align*} c_{t_0}=-\sum_{t\in (T\setminus S)\setminus\{t_0\}}c_t. \end{align*} Applying the non-Archimedean valuation inequality to the finite sum on the right gives $\nu(c_{t_0})\ge \min_{t\in (T\setminus S)\setminus\{t_0\}}\nu(c_t)$, contradicting the strict inequality above. Hence the minimum is attained at least twice.[/step]

[guided]We now explain why the valuation of the ordinary Plücker relation obtained in the previous step is exactly the tropical Plücker condition. In that step, the vector identity among the columns indexed by $T$ was wedged with the exterior product of the columns indexed by $S$, and extracting the coefficient of the basis element $\omega=e_1\wedge\cdots\wedge e_r$ gave signs $\sigma_t\in\{1,-1\}$ satisfying \begin{align*} \sum_{t\in T\setminus S}\sigma_t \Delta(S\cup\{t\})\Delta(T\setminus\{t\})=0. \end{align*} For every $t\in T\setminus S$, define the field element \begin{align*} c_t:=\sigma_t \Delta(S\cup\{t\})\Delta(T\setminus\{t\}) \in K. \end{align*} With this notation, the Plücker relation is the zero-sum identity \begin{align*} \sum_{t\in T\setminus S}c_t=0. \end{align*} The signs $\sigma_t$ do not affect valuations, because each $\sigma_t$ is either $1$ or $-1$, hence is a nonzero element of $K$ with valuation $0$. Using multiplicativity of the valuation, we get \begin{align*} \nu(c_t)=\nu(\Delta(S\cup\{t\}))+\nu(\Delta(T\setminus\{t\})). \end{align*} By the definition of $p$, this becomes \begin{align*} \nu(c_t)=p(S\cup\{t\})+p(T\setminus\{t\}). \end{align*} The key valuation fact is that a finite sum equal to zero cannot have a unique term of strictly smallest valuation. To see this in the present notation, suppose that one index $t_0\in T\setminus S$ uniquely minimizes $\nu(c_t)$. Then $c_{t_0}\ne 0$, and the zero-sum identity can be rewritten as \begin{align*} c_{t_0}=-\sum_{t\in (T\setminus S)\setminus\{t_0\}}c_t. \end{align*} The non-Archimedean valuation inequality says that the valuation of a finite sum is at least the minimum of the valuations of its summands. Applying it to the finite sum on the right-hand side gives $\nu(c_{t_0})\ge \min_{t\in (T\setminus S)\setminus\{t_0\}}\nu(c_t)$. But this contradicts the assumption that $\nu(c_{t_0})$ is strictly smaller than every other $\nu(c_t)$. Therefore the minimum among the values $\nu(c_t)$ is attained at least twice, which is exactly the assertion that \begin{align*} \min_{t\in T\setminus S}\bigl(p(S\cup\{t\})+p(T\setminus\{t\})\bigr) \end{align*} is attained for at least two distinct indices $t\in T\setminus S$.[/guided]

[step:Conclude that the valuation vector is a valuated matroid] The set $S\subset[n]$ of size $r-1$ and the set $T\subset[n]$ of size $r+1$ were arbitrary. Therefore $p$ satisfies every tropical Plücker relation of rank $r$ on $[n]$. Together with the fact that $p$ is not identically $\infty$, this is precisely the definition of a valuated matroid of rank $r$ on $[n]$. Hence the valuations of the maximal minors of $A$ form a valuated matroid. [/step]