Matrix Inverse is a Uniformly Continuous function for Uniformly Positive Definite matrices? Definition 1: A collection of $k\times k$ positive semidefinite matrices $\{A_n\}$ is said to be uniformly positive definite if for some $\eta > 0$, $det(A_n) > \eta$.
Definition 2: A function $f$ is uniformly continuous on $B$ if for any $\epsilon > 0$, there exists a $\delta > 0$ such that for $x, y \in B$, $d(x,y) < \delta$ implies $d(f(x),f(y)) < \epsilon$.
I need to prove that the matrix inverse is a uniformly continuous function for uniformly positive definite matrices, using the sup-metric, $d(M, N) = \max_{ij}|m_{ij} - n_{ij}| = ||M - N||_{\infty}$.
The statement is clearly true for scalars $\{a_n\}$, in that if we have $a_n > \eta > 0$ for all $n$, then have $m = \eta^{-1}$ larger than any other value of the function. For any $\epsilon >0$, let $\epsilon^* = \min(\epsilon, m/2)$. We take a $\delta$ such that $0 < \delta < (m - \epsilon^*)^{-1}$, and for that $\delta$ we have $|x - y| < \delta$ implies $|x^{-1} - y^{-1}| < \epsilon$.
However, I do not now how to generalize this proof to matrices. Any hint?
 A: The answer is NO.
Consider the matrix 
$$ A = \left(
\begin{matrix} a & c \\
c & a
\end{matrix} \right),
$$
where $c = a - a^{-1/2}$, and $a>0$ is large. With such choice for $a$ and $c$, the matrix $A$ is in the required class, as $A$ is positive definite and $det A = a^2 - (a-a^{-1/2})^2 = 2a^{1/2}  - a^{-1} >\eta$ for $a>0$ large.
For $\varepsilon > 0$ small consider the matrix 
$$ A_{\varepsilon} = \left(
\begin{matrix} a + \varepsilon & c - \varepsilon \\
c - \varepsilon & a + \varepsilon
\end{matrix} \right).
$$
Clearly $||A - A_\varepsilon||_{\infty} = \varepsilon$ and $A_\varepsilon$ is also in the required class.
We have
$$
A^{-1} = \frac{1}{a^2 - c^2} \left(
\begin{matrix} a & - c \\
- c & a
\end{matrix} \right)  \ \text{ and }  \ A_{\varepsilon}^{-1} = \frac{1}{(a + \varepsilon)^2 - (c - \varepsilon)^2} \left(
\begin{matrix} a + \varepsilon & -(c - \varepsilon) \\
-( c - \varepsilon) & a + \varepsilon
\end{matrix} \right).
$$
For the first elements of these inverses and with the choice of $c$ as above, we get
\begin{equation}
\frac{a}{a^2 - c^2} - \frac{a + \varepsilon}{(a+\varepsilon)^2 - (c - \varepsilon)^2} = \frac{a}{2 a^{1/2} - a^{-1} } - \frac{a + \varepsilon}{ (2a - a^{-1/2})(2 \varepsilon + a^{-1/2}) }. \tag{1}
\end{equation}
Since $||A - A_\varepsilon||_{\infty} $ is independent of $a$, for uniform continuity of the inverses we need to make $(1)$ small for any $a>0$ large. But for $\varepsilon >0$ fixed, the limit of $(1)$ as $ a \to + \infty $ is infinite, meaning that the uniform continuity in question fails.
