Before phrasing my question, I have to mention some definitions and theorems to make sure we stay on the same ground. However, it may become a little bit lengthy and you can take a look at the end of this post to see the main question. So, let's get started.
Canonical Form
Let us consider a general linear programming problem of the following canonical form
\begin{align} \min & \quad \mathbf{c}^\top \mathbf{x} \\ \text{subject to} & \quad \mathsf{G} \, \mathbf{x} \ge\mathbf{f}, \end{align}
where $\mathbf{c}, \, \mathbf{x} \in \mathbb{R}^n$, $\mathbf{f} \in \mathbb{R}^l$ are column vectors and $\mathsf{G}\in\mathbb{R}^{l,\, n}$ is a matrix with $l$ rows and $n$ columns. Let $P = \{\mathbf{x}\in\mathbb{R}^n | \,\mathsf{G} \, \mathbf{x} \ge \mathbf{f} \}$ be the corresponding feasible set which is called a polyheron and is known to be a convex set. Based on our geometric intuition, we have the following definition for a basic solution of $P$.
Definition 1. Basic Solution. The point $\mathbf{y}$ is called a basic solution if there are $n$ linearly independent active constraints at $\mathbf{y}$. More precisely, this means that there is an index set $I\subset\{1,\dots,l\}$ such that $|I|=n$, $\{\mathbf{g}_i \, | \,\mathbf{g}_i^\top = \mathsf{G}_{[i,\,:]},\, i\in I\}$ is linearly independent, and $\mathbf{g}_i^{\top}\mathbf{y} = f_i$ for every $i\in I$. Clearly, $n\leq l$ is a necessary condition for the existence of a basic solution.
Standard Form
Now, let us pay attention to the standard form of linear programming problem that reads as \begin{align} \min & \quad \mathbf{c}^\top \mathbf{x} \\ \text{subject to} & \quad \mathsf{A} \, \mathbf{x} = \mathbf{b}, \\ & \quad \mathbf{x} \ge \mathbf{0} \end{align}
where $\mathsf{A}\in\mathbb{R}^{m,\, n}$ is a matrix with $m$ rows and $n$ columns and is a full row-rank matrix that is $\text{rank}\,\mathsf{A} = m \leq n$. Let $S = \{\mathbf{x}\in\mathbb{R}^n | \,\mathsf{A} \, \mathbf{x} = \mathbf{b}, \, \mathbf{x} \ge \mathbf{0} \}$ be the corresponding feasible set, which is called a standard polyhedron. In comparison with the canonical form, we observe that $\mathsf{G} = [\mathsf{A}^\top|-\mathsf{A}^\top|\,\mathsf{I}\,]^\top$ , $\mathbf{f}=[\mathbf{b}^\top|\mathbf{0}]^\top$, and $l=2m+n$.
Definition 2. Standard Basic Solution. Let $\mathbf{y}\in S$ be a basic solution. If $\mathsf{A}\mathbf{y}=\mathbf{b}$ then $\mathbf{y}$ is said to be a standard basic solution.
Definition 3. Basis. Let $\beta = \{\beta_1,\dots, \beta_m\}\subset \{1,\dots, n\}$ be an index set. Moreover, let $\bar{\beta}=\{\bar{\beta}_1,\dots,\bar{\beta}_{n-m}\}=\{1,\dots, n\}-\beta$ be its complement. The index sets $\beta$ and $\bar{\beta}$ induce a natural partition on $m$ by $n$ matrices and vectors of length $n$. More specifically, they partition $\mathsf{A}$ into the matrices $\mathsf{B} = [\mathbf{a}_{\beta_1},\dots,\mathbf{a}_{\beta_m}]$ and $\mathsf{N}=[\mathbf{a}_{\bar{\beta}_1},\dots,\mathbf{a}_{\bar{\beta}_{n-m}}]$ and every $\mathbf{x}\in \mathbb{R}^n$ into $\mathbf{x}_{\beta}=[x_{\beta_1},\dots,x_{\beta_n}]^\top$ and $\mathbf{x}_{\bar{\beta}} = [x_{\bar{\beta_1}},\dots,x_{\bar{\beta}_{n-m}}]^\top$. Let $\mathbf{y}\in S$ be a standard basic solution. We say that $\beta$ is a basis associated with $\mathbf{y}$ if $\{\mathbf{a}_i|i\in\beta\}$ is linearly independent (or equivalently $\mathsf{B}$ is invertible) and $\mathbf{y}_{\bar\beta} = \mathbf{0}$. In this case, $\mathsf{B}$ is called the basis matrix, $\mathbf{y}_\beta$ are said to be basic variables and $\mathbf{y}_{\bar{\beta}}$ are named non-basic variables.
Then, the following theorem suggests a way for obtaining standard basic solutions of $S$. Indeed, it connects our geometric view about basic solutions to an algebraic one that simplifies their computation.
Theorem 1. Let $\mathbf{y}\in \mathbb{R}^n$ be such that $\mathsf{A}\mathbf{y}=\mathbf{b}$. Then, $\mathbf{y}$ is a standard basic solution of $S$ if and only if there exists a basis $\beta$ associated with $\mathbf{y}$. Furthermore, we have $\mathbf{y}_{\beta} = \mathsf{B}^{-1}\mathbf{b}$. As a consequence, for every standard basic solution there is a basis and every basis uniquely determines a standard basic solution.
Adjacency
Based on our geometric intuition, one can suggest the following definition for adjacency of two basic solutions.
Definition 4. Geometric Adjacency. Let $\mathbf{u} \ne \mathbf{v}$ be two distinct basic solutions of $P$. They are said to be geometrically adjacent if there are $n-1$ linearly independent constraints which are active at both of them. More precisely, this means that there is an index set $I\subset\{1,\dots,l\}$ such that $|I|=n-1$, $\{\mathbf{g}_i \, | \,\mathbf{g}_i^\top = \mathsf{G}_{[i,\,:]},\, i\in I\}$ is linearly independent, and $\mathbf{g}_i^{\top}\mathbf{u} = \mathbf{g}_i^{\top}\mathbf{v} = f_i$ for every $i\in I$.
Also, Theorem 1 gives rise to the following algebraic definition of adjacency for two standard basic solutions.
Definition 5. Algebraic Adjacency. Let $\mathbf{u} \ne \mathbf{v}$ be two distinct standard basic solutions of $S$. They are said to be algebraically adjacent if there are two corresponding basis $\beta$ and $\gamma$ that differ in only one index. Furthermore, $\mathbf{u}_{\bar{\beta}} = \mathbf{0}$ and $\mathbf{v}_{\bar{\gamma}} = \mathbf{0}$.
Question
Prove the following equivalency.
Theorem 2. Adjacent Standard Basic Solutions. Let $\mathbf{u} \ne \mathbf{v}$ be two distinct standard basic solutions of $S$. Then, $\mathbf{u}$ and $\mathbf{v}$ are geometrically adjacent if and only if they are algebraically adjacent.
Let $\mathbf{u}$ and $\mathbf{v}$ be two distinct algebraically adjacent standard basic solutions. According to Theorem 1, there are bases $\beta$ and $\gamma$ associated with these basic solutions. Furthermore, since they are adjacent, they only differ in one index. Let $\beta=\{\beta_1,\dots,\beta_p,\dots,\beta_m\}$ be an associated basis with $\mathbf{u}$ and suppose that when $k\in\bar{\beta}=\{1,\dots,n\}-\beta$ is interchanged with $\beta_p$ we obtain a basis $\gamma = \{\beta_1,\dots,k,\dots,\beta_m\}$ associated with $\mathbf{v}$. According to the definition of standard basic solution, both $\mathbf{u}$ and $\mathbf{v}$ must satisfy the following $n-1$ number of equations
\begin{align} \mathsf{A} \, \mathbf{x} &= \mathbf{b} \\ x_i &= 0, \qquad i \in \bar{\beta} \cap \bar{\gamma}, \end{align}
where $|\bar{\beta} \cap \bar{\gamma}| = n-m-1$. In order to construct the coefficient matrix $\mathsf{K}$ of the above system, let us make some definitions. Let $\beta_{m+1}=k$ and $\bar{\beta} \cap \bar{\gamma} = \{\beta_{m+2},\dots,\beta_{n}\}$. Define $\mathsf{E}^\top =[\mathbf{E}_{\beta_{m+2}},\dots,\mathbf{E}_{\beta_{n}}]$ where $\mathbf{E}_i$ is the $i$th standard unit vector in $\mathbb{R}^n$, $\mathbf{e}_i$ to be the $i$th unit vector in $\mathbb{R}^{n-m-1}$, $\mathbf{A}_i$ as the $i$th row and $\mathbf{a}_i$ as the $i$th column of $\mathsf{A}$. Now, notice that
\begin{align} \mathsf{K}\,\mathsf{P} = \begin{bmatrix} \mathsf{A} \\ \mathsf{E} \end{bmatrix} \mathsf{P} &= \begin{bmatrix} \mathbf{A}_1 & \dots & \mathbf{A}_m & \mathbf{E}_{\beta_{m+2}} & \dots & \mathbf{E}_{\beta_n} \end{bmatrix}^\top \begin{bmatrix} \mathbf{E}_{\beta_1} & \dots & \mathbf{E}_{\beta_n} \end{bmatrix} \\ &= \begin{bmatrix} \mathbf{a}_{\beta_{1}} & \dots & \mathbf{a}_{\beta_{m}} & \mathbf{a}_k &\mathbf{a}_{\beta_{m+2}} & \dots & \mathbf{a}_{\beta_n} \\ \mathbf{0} & \dots & \mathbf{0} & \mathbf{0} & \mathbf{e}_1 & \dots & \mathbf{e}_{n-m-1} \end{bmatrix} \\ &= \begin{bmatrix} \mathsf{B} & \mathbf{a}_k & \mathsf{M} \\ \mathsf{O} & \mathbf{0} & \mathsf{I} \end{bmatrix} \\ &= \mathsf{L}, \end{align}
where $\mathsf{P}$ is a permutation matrix that interchanges the columns of $\mathsf{K}$ such that the structure shown can be observed. Since $\mathsf{P}$ is invertible we have
\begin{equation} \text{rank}\,\mathsf{K} = \text{rank}\,\mathsf{K}\,\mathsf{P} = \text{rank}\, \mathsf{L} = n-1, \end{equation}
which verifies that $\mathsf{K}$ has $n-1$ linearly independent rows. Consequently, there are $n-1$ linearly independent constraints that are active at both $\mathbf{u}$ and $\mathbf{v}$, proving that they are geometrically adjacent.
Now, assume that $\mathbf{u}$ and $\mathbf{v}$ are geometrically adjacent. By definition of geometric adjacency, both of them must satisfy the following $n-1$ linearly independent constraints
\begin{align} \mathsf{A} \, \mathbf{x} &= \mathbf{b} \\ x_i &= 0, \qquad i \in \bar{\alpha}, \end{align}
where $|\bar{\alpha}|=n-m-1$. Since $\mathbf{u}$ and $\mathbf{v}$ are standard basic solutions, there must exist two distinct indices $p,q\in\alpha$ such that they satisfy the following $n$ linearly independent constraints
\begin{alignat}{2} \mathsf{A} \, \mathbf{u} &= \mathbf{b} && \mathsf{A} \, \mathbf{v} &= \mathbf{b} \\ u_i &= 0, \qquad i \in \bar{\alpha}\cup\{p\}, \qquad\qquad && v_i &= 0, \qquad i \in \bar{\alpha}\cup\{q\}. \end{alignat}
Take $\beta=\alpha-\{p\}$ and $\gamma=\alpha-\{q\}$ as the candidates for the associated basis with $\mathbf{u}$ and $\mathbf{v}$, respectively. It is obvious that $\mathbf{u}_{\bar\beta}=\mathbf{0}$ and $\mathbf{v}_{\bar\gamma}=\mathbf{0}$. Furthermore, since the above linear systems have unique solutions, substituting the second set of equations into the first one implies that the following linear systems
\begin{align} \sum_{i\in\beta} u_i\mathbf{a}_i = \mathbf{b}, \qquad \qquad \qquad \sum_{i\in\gamma} v_i\mathbf{a}_i = \mathbf{b}, \end{align}
must have a unique solution, which in turn results in linear independence of $\{\mathbf{a}_i|i\in\beta\}$ and $\{\mathbf{a}_i|i\in\gamma\}$. Since $\beta$ and $\gamma$ only differ in one index, we conclude that $\mathbf{u}$ and $\mathbf{v}$ are algebraically adjacent. This completes the proof.