Why is the set of $n\times n$ real, non-singular matrices an open subset of the set of all $n\times n$ real matrices? I don’t quite understand what ”open” means in this context.
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"Open" is in the sense of topology. A set can be endowed with an extra structure, called a topology, that is a decision about which subsets of our set will be called "open" (the decision cannot be arbitrary however, there are some rules). The real numbers, $\mathbb{R}$, can be given a topology, called the "usual topology", where we call $S\subseteq \mathbb{R}$ "open" when, for any $x\in S$, there is an open interval $(a,b)$ such that $x\in (a,b)$ and $(a,b)\subseteq S$.
The set $M_n(\mathbb{R})$ of all $n\times n$ real matrices can be identified with $\mathbb{R}^{n^2}$, via $$\begin{pmatrix}a_{11} & \cdots & a_{1n}\\ \vdots& \ddots & \vdots\\ a_{n1}& \cdots& a_{nn}\end{pmatrix}\mapsto(a_{11},a_{12},\ldots,a_{1n},a_{21},\ldots,a_{nn})$$ and because $\mathbb{R}$ has its usual topology, we can put the product topology on $\mathbb{R}^{n^2}$, and hence on $M_n(\mathbb{R})$.
The determinant function $\det:M_n(\mathbb{R})\to\mathbb{R}$ sends an $n\times n$ matrix to its determinant. It is continuous, because it is, in fact, a polynomial in the entries of the matrix (i.e., under our identification, it is a polynomial function in $n^2$ variables from $\mathbb{R}^{n^2}$ to $\mathbb{R}$). One of the properties of continuous functions (this is in fact what the definition of a continuous function is in general) is that the preimage of an open set is open. Then because a matrix is non-singular if and only if its determinant is not $0$, we have that $$\text{GL}_n(\mathbb{R})=\{A\in M_n(\mathbb{R})\mid A\text{ is non-singular}\}={\det}^{-1}(\mathbb{R}-\{0\})$$ and $\mathbb{R}-\{0\}$ is an open set of $\mathbb{R}$, hence $\text{GL}_n(\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$.
The notation $\text{GL}$ comes from the fact that this is often called the general linear group.
It's also interesting to contrast this result with another result, which I showed in a different answer, that the set of $m\times n$ matrices having rank less than or equal to $k$, for some $k\leq\min\{m,n\}$, is a closed subset of $M_{m\times n}(\mathbb{R})$. Note that a square $n\times n$ matrix is non-singular if and only if it has full rank, i.e. its rank is equal to $n$.
Zev Chonoles
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Somewhat informally: if you take a matrix, and change the entries of it a little bit, then the determinant also changes by a little bit. (Formally, the determinant is a continuous function of the matrix entries.)
So if you start with a matrix with non-zero determinant, and change its entries a little bit, the result will also have non-zero determinant. So matrices that are "near" a non-singular matrix are themselves non-singular; that's what it means for that the set of non-singular matrices is an open set.
In contrast, the set of singular matrices is not open. Since it's the complement of an open set, it's called a closed set: Wikipedia article on closed sets. It's not open because if you take a non-singular matrix and change it a little bit, it (usually) won't be non-singular any more; it's closed because if you take the limit of a sequence of singular matrices (in the "obvious" componentwise way) and that limit exists, it is a singular matrix.
(All this is made more formal in Zev Chonoles' answer.)
Michael Lugo
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Try to prove this: if $A$ is an invertible matrix & $\|B-A\|<\|A^{-1}\|$, then $B$ is an invertible matrix.
Rócherz
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A result is that $\|B\| < 1 \implies I+B$ is invertible, since for any $x$ such that $\|x\| = 1$, $${\|(I+B)x\|} \ge {\|Ix\| - \|Bx\|} \ge {1 - \|B\|} > {0.}$$
So a unit ball around the identity lies in the space of all invertible matrices.
We can easily extend this to any invertible matrix. If $\|\delta A\| < \frac1{\|A^{-1}\|}$ then $A+\delta A$ is invertible. $${\|A^{-1}\delta A\| \le \|\delta A\|\ \|A^{-1}\| < 1} \implies {(I + A^{-1}\delta A)\text{ is invertible}}\implies {A + \delta A\text{ is invertible}.}$$
So every invertible matrix can be perturbed slightly, in any direction, and we still end up with an invertible matrix. This is exactly what it means for that set to be open.
