Dimension of the intersection of $k > n$ hyperplanes Suppose there is an $n$-dimensional affine subspace - let us call it $S$ of
$\mathbb{R}^d$, $d>n$. Suppose we have $k>n$ hyperplanes in $\mathbb{R}^d$. We want to determine the dimensions of the intersection of these hyperplanes and $S$. It is given that the intersection is non-empty.
A very helpful answer here says, for $k>n$ the intersection can have any number of dimensions from $0$ to $n−1$. My question is: How do we determine the actual dimensions of the intersection? My hyperplanes are all of the form $x_{l} = x_{l'}$, i.e. they stipulate equality of pairs of coordinates of a point in $d$ dimensions.
My approach: The $k$ equalities are all independent. But it is also given that the intersection is non-empty. I'm at a loss trying to determine how this is possible.
Edit: An earlier version of the question did not metion $\mathbb{R}^d$ at all, reduced the space to $S$ and spoke as if we have a set of $k>n$ equalities, which create a subspace of $S$, which in turn implies that some of the $k$ equalities (when the space is reduced to S) have to be redundant (since it is given that the intersection is non-empty). I subsequently edited the question to add the details about $\mathbb{R}^d$ at the suggestion of user Zanxiong, who very helpfully provided the first answer.

Express each hyperplane by its analytical equation $a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n = b_i, i = 1, \ldots, k$. Then the intersection is the solution to the linear system $Ax = b$, i.e., the set $S = \{x \in \mathbb{R}^n: Ax = b\}$, where
\begin{align*}
A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{k1} & a_{k2} & \cdots & a_{kn} 
\end{pmatrix}, \quad
x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \quad
b = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix}.
\end{align*}
Now you can unleash the theory of linear system to address the problem. When $b \neq 0$, then $S \neq \emptyset$ if and only if $\text{rank}((A, b)) = \text{rank}(A)$. However, in general $S$ is not a subspace of $\mathbb{R}^n$, hence it's pointless to discuss the dimension of $S$.
When $b = 0$, then it is well known that $\dim(S) = n - \text{rank}(A)$. So to determine the dimension of $S$ boils down to determine the rank of the coefficient matrix $A$. Since $0 \leq \text{rank}(A) \leq \min(n, k) = n$, $0 \leq \dim(A) \leq n$.
