 Solution

Combine all inequalities, including nonnegativity, into $Cx\le d$; explicitly, the rows $-e_j^{T}x\le 0$ encode $x_j\ge 0$. Let $x_{\ast }$ be a vertex of a linear polyhedron, equivalently an extreme point of the feasible linear polyhedron, and let $I$ index the active constraints there.

Suppose the active row normals have rank smaller than $n$. By the rank-nullity theorem there is a nonzero vector $h$ with $C_{i}h=0$ for every $i\in I$. Every inactive row has positive slack variable value $s_i=d_i-C_i{x}_{\ast }>0$. Since there are only finitely many rows, choose $\epsilon >0$ so small thatThen both $x_{\ast }+\epsilon h$ and $x_{\ast }-\epsilon h$ are feasible points: the active constraints remain equalities and no inactive inequality is violated. They are distinct, and $x_{\ast }$ is their midpoint, contradicting the extreme point property.

Therefore the active row normals have rank $n$, and at least $n$ linearly independent constraints are satisfied with equality. In particular at least $n$ constraints are tight, as claimed. Nonnegativity bounds must be counted: for example, at the origin these can supply all $n$ active rows. Degeneracy can make more than $n$ constraints tight, so the result does not assert exactly $n$.