# 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?

• What metric are you using on the space of matrices? May 29, 2018 at 5:45
• @Merudo, if you use the constructive definition of the inverse, that for an $n\times n$ invertible matrix $A$, the inverse equals $\frac{1}{det A} adj(A)$, where $adj(A)$ is the adjugate matrix of $A$ (see Wikipedia), then all components of $A^{-1}$ can easily be controlled by the elements of the original matrix $A$. You simply need to utilize how the inverse is build from the original matrix, then continuity properties of $A^{-1}$ will be delegated to the elements of $A$. The metric you use, doesn't matter that much, we're in a finite dimensional space.
– Hayk
May 29, 2018 at 5:57
• @Hayk thank you for the hint. If I understand right, I should first prove that the $1/det(A)$ and $adj(A)$ are both uniformly continuous functions? May 29, 2018 at 6:20
• @Merudo, yes, you can do that, but note that $adj(A)$ is a matrix, and you need to show continuity of its components with respect to $A$. For instance, how you show continuity of the determinant? You observe, that the determinant is a finite sum of finite products of elements of $A$ (with signs). All operations you perform to compute the determinant, are continuous.
– Hayk
May 29, 2018 at 6:37
• @Hayk true, but continuity is not enough, we need uniform continuity. May 29, 2018 at 6:40

## 1 Answer

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 $$\frac{a}{a^2 - c^2} - \frac{a + \varepsilon}{(a+\varepsilon)^2 - (c - \varepsilon)^2} = \frac{a}{2 a^{1/2} - a^{-1} } - \frac{a + \varepsilon}{ (2a - a^{-1/2})(2 \varepsilon + a^{-1/2}) }. \tag{1}$$

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.