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 set
$$
U_I=\{M:\det M_I\ne0\}
$$
is <open set>[open] in the <vector space> of all real $m$ by $n$ matrices, because the determinant is <continuous function>[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 of a vector space>[dimension] $mn$. This is the <full-row-rank matrix manifold>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 1 / a / Solution

{title2=$X_{m,n}$}

The real $m$ by $n$ matrices of <matrix rank> $m\leq n$ form an open <smooth manifold> $X_{m,n}\subseteq\mathbb R^{mn}$. Indeed, it is the union of the open sets on which a chosen $m$ by $m$ minor has nonzero <determinant>, so its dimension is $mn$.

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