# Matrices of Full Rank is a Smooth Manifold Proof Clarification

In Lee's Introduction to Smooth Manifolds, he proves that matrices of full rank are a manifold as such:

Suppose $m < n$, and let $M_m(m \times n, \mathbb{R})$ denote the subset of $M(m \times n, \mathbb{R})$ consisting of matrices of rank $m$. If $A$ is an arbitrary such matrix, the fact that rank $A = m$ means that $A$ has some nonsingular $m \times m$ minor. By continuity of the determinant function, this same minor has nonzero determinant on some neighborhood of $A$ in $M(m \times n, \mathbb{R})$, which implies that A has a neighborhood contained in $M_m(m \times n, \mathbb{R})$.

Bolding my own, as that's the part I'm having a little trouble with. My understanding is that, because the matrix is invertible it is contained in $\text{det}^{-1}((-\infty, 0) \cup (0, \infty)) \subseteq GL(m, \mathbb{R})$, which is open, since $\text{det}$ is continuous. Then, as $GL(m, \mathbb{R})$ has the subspace topology inherited by $M(m, \mathbb{R})$, this must be contained in an open set of $M(m, \mathbb{R})$, which is in turn contained in an open subset of $M(m \times n, \mathbb{R})$. Then, the intersection of this set with $M_m(m \times n, \mathbb{R})$ will be a neighbourhood $A$. I have two questions:

1. Do I have the right understanding of this argument?
2. How do we know that the vector space topology that on $M(m, \mathbb{R})$ agrees with the subspace topology it will inherit from $M(m \times n, \mathbb{R})$?

You do understand the argument correctly. I think one could make it simpler by picking $A\in M(m\times n,\mathbb{R})$ and considering the function $\det_m:M(m\times n,\mathbb{R})\rightarrow \mathbb{R}$ which takes the determinant of the $m\times m$ minor which is nonsingular at $A$. This is continuous since it is polynomial in the matrix entries. Now we can find an open set $U$ such that $${\det}_m (A)\in U\subset \mathbb{R}\setminus\{0\}.$$ Then $\det_m^{-1}(U)\subset M_m(m\times n,\mathbb{R})$ is an open neighborhood of $A$.
To your second question: I am not sure what $\textit{the}$ vector space topology on $M(m,\mathbb{R})$ means exactly. Maybe it is something I am unaware of, but as I understand it, generally, $M(m\times n,\mathbb{R})$ is the product of $mn$ copies of $\mathbb{R}$ and would be given the product topology, from the standard topology on $\mathbb{R}$. One could also use the inner product of $\mathbb{R}^{mn}\cong M(m\times n,\mathbb{R})$ to define a metric topology, which turns out to be the same as the product topology. Then the topology of $M(m,\mathbb{R})$ as a product of $\mathbb{R}$'s is the same as its topology as a subset of $M(m\times n,\mathbb{R})$. (I don't know if any of this was unknown to you but I thought I would write it in case it was.)
• As to "the" topology on $M(m, \mathbb{R})$, I mean essentially what you mean by the inner product: any norm on a real, finite dimensional vector space will induce the same topology, and at least one norm exists. Of course there are other topologies, but that's the one I was referring to. I suppose what my question boils down to is, if $m < n$ how do we know that the product topology on $\mathbb{R}^m$ is the same as the subspace topology it inherits from $\mathbb{R}^n$? – Duncan Ramage May 30 '17 at 19:10