Neighborhood of the identity matrix

Question

Is it correct to say that there exists a neighborhood of the identity matrix which is a subset of the set of nonsingular matrices, because the determinant is continuous over the set of matrices?

Answer 1

Yes, this is true. The determinant of a matrix $A$ can actually be computed as a linear combination of products of the entries, and so the determinant is a continuous function of the entries. The identity matrix has a determinant of $1$, so if we consider the the determinant as a function $D:\mathbb{R}^{n\times n}\to\mathbb{R}$, the inverse image of the interval $(0,2)$ is an open set in $\mathbb{R}^{n\times n}$ containing the identity matrix. Thus, this set is a neighborhood of $I$, and all matrices in this set are invertible.

Answer 2

Yes! This is true for exactly the reason you describe!

Answer 3

Yes!!!

Let

$\Bbb F = \Bbb R \; \text{or} \; \Bbb C. \tag 0$

Since

$\det: M_n(\Bbb F) \to \Bbb F, \; A \to \det(A) \tag 1$

is continuous, which follows from the fact that it may be expressed as a polynomial in the entries of $A$, and since

$\det I = 1, \tag 2$

it follows that for any open disk about $1$ of the form

$D_\epsilon(1) = \{ x \in \Bbb F \mid \vert x - 1 \vert < \epsilon, \; 0 < \epsilon < 1\}, \tag 3$

we have

$\det^{-1}(D_\epsilon(1)) \subset M_n(\Bbb F) \tag 4$

is open; furthermore,

$I \in \det^{-1}(D_\epsilon(1)), \tag 5$

by virtue of (2). Now it is easy to see that

$A \in \det^{-1}(D_\epsilon(1)) \Longrightarrow \det A \ne 0 \notin D_\epsilon(1) \tag 6$

since $\epsilon < 1$.

We thus see that $\det^{-1}(D_\epsilon(1))$ is such a desired open set.