 Solution

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.