Proof of Perturbations of an Invertible Matrix: the Invertible Matrices Form an Open Set and Inversion Is Continuous

Each result cited is universally quantified over the data in its own statement.

The clauses perturbation and continuous are proved for an arbitrary invertible $A\in\mathcal{M}_{n}(\mathbb{R})$. The clause open applies perturbation with $C$, and continuous-map applies continuous with $C$, in place of $A$. The Euclidean norm on $\mathbb{R}^{n}$ is that of The Euclidean Norm on $\mathbb{R}^n$ §norm, also written $\lVert\cdot\rVert$. For an invertible $C\in\mathcal{M}_{n}(\mathbb{R})$, $CC^{-1}=I_{n}$ and $C^{-1}C=I_{n}$, by Inverse Matrix and Invertible Real Square Matrix §invertible and Inverse Matrix and Invertible Real Square Matrix §inverse, with $I_{n}$ the identity matrix. By The Distance between Real Matrices §distance, $d_{n}(C,D)=\lVert C-D\rVert$ for $C,D\in\mathcal{M}_{n}(\mathbb{R})$, and $d_{n}(C,D)\ge0$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §nonnegative.

Clause perturbation. Let $B\in\mathcal{M}_{n}(\mathbb{R})$, and put $\alpha=\lVert A^{-1}\rVert$ and $\beta=d_{n}(A,B)$. Then $\alpha\ge0$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §nonnegative, $\beta\ge0$, and we assume $\alpha\beta<1$, so $1-\alpha\beta>0$.

Step 1: an estimate through $\lVert Bx\rVert$. Let $x\in\mathbb{R}^{n}$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $x=I_{n}x=(A^{-1}A)x=A^{-1}(Ax)$, so $\lVert x\rVert\le\alpha\,\lVert Ax\rVert$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §bound. Since $Ax=Bx+(Ax-Bx)$, Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §triangle and The Real Matrices with the Operator-Norm Distance Form a Complete Metric Space with Entrywise Convergence §vector-bound give $\lVert Ax\rVert\le\lVert Bx\rVert+\lVert Ax-Bx\rVert\le\lVert Bx\rVert+\beta\lVert x\rVert$. Multiplying by $\alpha\ge0$, $\lVert x\rVert\le\alpha\lVert Bx\rVert+\alpha\beta\lVert x\rVert$, that is,

$$(1-\alpha\beta)\,\lVert x\rVert\le\alpha\,\lVert Bx\rVert\qquad(x\in\mathbb{R}^{n}).\tag{1}$$

Step 2: $B$ is invertible. Let $x\in\mathbb{R}^{n}$ with $Bx=0$. Then $\lVert Bx\rVert=0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §vanishing, so $(1-\alpha\beta)\lVert x\rVert\le0$ by (1); as $1-\alpha\beta>0$ and $\lVert x\rVert\ge0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square, $\lVert x\rVert=0$, and $x=0$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §vanishing. So $B$ is invertible, by Criteria for Invertibility of a Square Real Matrix §trivial-kernel and Criteria for Invertibility of a Square Real Matrix §invertible.

Step 3: a bound for $\lVert B^{-1}\rVert$. Let $y\in\mathbb{R}^{n}$ and $x=B^{-1}y$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $Bx=B(B^{-1}y)=(BB^{-1})y=I_{n}y=y$, so (1) gives $(1-\alpha\beta)\lVert B^{-1}y\rVert\le\alpha\lVert y\rVert$, that is, $\lVert B^{-1}y\rVert\le\frac{\alpha}{1-\alpha\beta}\lVert y\rVert$. As this holds for every $y\in\mathbb{R}^{n}$, Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §least gives

$$\lVert B^{-1}\rVert\le\frac{\alpha}{1-\alpha\beta}.\tag{2}$$

Step 4: the distance between the inverses. Comparing entries by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality, $A-B=A+(-1)B$ and $B^{-1}+(-1)A^{-1}=B^{-1}-A^{-1}$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §distributive, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §scalar, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §associative and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity, with $AA^{-1}=I_{n}$ and $B^{-1}B=I_{n}$,

$$\bigl(B^{-1}(A-B)\bigr)A^{-1}=\bigl(B^{-1}A+(-1)(B^{-1}B)\bigr)A^{-1}=B^{-1}(AA^{-1})+(-1)(I_{n}A^{-1})=B^{-1}+(-1)A^{-1}=B^{-1}-A^{-1}.$$

By The Real Matrices with the Operator-Norm Distance Form a Complete Metric Space with Entrywise Convergence §metric (Metric Space §symmetric), Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products §product applied twice, the nonnegativity of $\alpha$ and $\beta$, and (2),

$$d_{n}(A^{-1},B^{-1})=d_{n}(B^{-1},A^{-1})=\bigl\lVert\bigl(B^{-1}(A-B)\bigr)A^{-1}\bigr\rVert\le\lVert B^{-1}\rVert\,\lVert A-B\rVert\,\lVert A^{-1}\rVert\le\frac{\alpha}{1-\alpha\beta}\,\beta\,\alpha=\frac{\alpha^{2}\beta}{1-\alpha\beta}.$$

Clause open. Let $C\in\mathrm{GL}_{n}(\mathbb{R})$, that is, $C\in\mathcal{M}_{n}(\mathbb{R})$ invertible ($\mathrm{GL}_{n}(\mathbb{R})\subseteq\mathcal{M}_{n}(\mathbb{R})$, with these elements, by The Invertible Matrices and the Positive Semidefinite Matrices Form Sets §invertible), and put $\gamma=\lVert C^{-1}\rVert\ge0$ and $r=1/(\gamma+1)>0$. Let $B$ be in the open ball $B_{d_{n}}(C,r)$, so that $B\in\mathcal{M}_{n}(\mathbb{R})$ and $d_{n}(C,B)<r$ by Open Ball in a Metric Space §membership. Since $d_{n}(C,B)\ge0$ and $\gamma\le\gamma+1$, $\gamma\,d_{n}(C,B)\le(\gamma+1)\,d_{n}(C,B)<(\gamma+1)\,r=1$. By the clause perturbation with $C$ in place of $A$, $B$ is invertible, so $B\in\mathrm{GL}_{n}(\mathbb{R})$. Hence $B_{d_{n}}(C,r)\subseteq\mathrm{GL}_{n}(\mathbb{R})$, and $\mathrm{GL}_{n}(\mathbb{R})$, a subset of $\mathcal{M}_{n}(\mathbb{R})$, is open by Open Subset of a Metric Space §open.

Clause continuous. Let $A\in\mathcal{M}_{n}(\mathbb{R})$ be invertible. The choices are made in this order: first $\alpha=\lVert A^{-1}\rVert\ge0$, fixed by $A$; then $\varepsilon>0$ is given; then

$$\delta=\min\Bigl\{\frac{1}{2(\alpha+1)},\ \frac{\varepsilon}{2(\alpha+1)^{2}}\Bigr\}>0$$

is chosen from $\alpha$ and $\varepsilon$ alone; finally $B\in\mathcal{M}_{n}(\mathbb{R})$ with $\beta=d_{n}(A,B)<\delta$ is arbitrary. Since $0\le\alpha\le\alpha+1$ and $\beta\ge0$,

$$\alpha\beta\le(\alpha+1)\beta<(\alpha+1)\delta\le\tfrac12<1 .$$

By the clause perturbation, $B$ is invertible and $d_{n}(A^{-1},B^{-1})\le\alpha^{2}\beta/(1-\alpha\beta)$. As $1-\alpha\beta>\frac12$ and $\alpha^{2}\beta\ge0$, and as $\alpha^{2}\le(\alpha+1)^{2}$,

$$d_{n}(A^{-1},B^{-1})\le\frac{\alpha^{2}\beta}{1-\alpha\beta}\le2\alpha^{2}\beta\le2(\alpha+1)^{2}\beta<2(\alpha+1)^{2}\delta\le\varepsilon .$$

So $\delta$ has the required property.

Clause continuous-map. For $B\in\mathrm{GL}_{n}(\mathbb{R})$, that is, $B\in\mathcal{M}_{n}(\mathbb{R})$ invertible, $B^{-1}$ is the only $C\in\mathcal{M}_{n}(\mathbb{R})$ with $BC=I_{n}$ and $CB=I_{n}$, by Inverse Matrix and Invertible Real Square Matrix §inverse and Inverse Matrix and Invertible Real Square Matrix §invertible. Let $\pi:\mathcal{M}_{n}(\mathbb{R})\times\mathcal{M}_{n}(\mathbb{R})\to\mathcal{M}_{n}(\mathbb{R})$ be the map giving the matrix product, $\pi((B,C))=BC$, a map by Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices §maps, so that $BC=I_{n}$ if and only if $((B,C),I_{n})\in\pi$. The condition on $C$ is then the formula $\theta(C,B)$ under its guard:

$$\theta(C,B):\ C\in\mathcal{M}_{n}(\mathbb{R})\wedge((B,C),I_{n})\in\pi\wedge((C,B),I_{n})\in\pi,\qquad\text{guard }B\in\mathrm{GL}_{n}(\mathbb{R}),$$

with $\mathcal{M}_{n}(\mathbb{R})$, $I_{n}$, $\pi$ and $\mathrm{GL}_{n}(\mathbb{R})$ as parameters; $\theta$ and the guard are predicative (Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices §predicativity), and under the guard exactly one set $C$ satisfies $\theta(C,B)$, namely $B^{-1}$; so $B^{-1}$ is a defined set symbol, used properly for $B\in\mathrm{GL}_{n}(\mathbb{R})$, with $B^{-1}\in\mathcal{M}_{n}(\mathbb{R})$. Hence, by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol §map-symbol with $k=0$, the expression $t(u)=u^{-1}$, and $\mathrm{GL}_{n}(\mathbb{R})$ and $\mathcal{M}_{n}(\mathbb{R})$ in place of its sets $A$ and $B$, there is a map $f:\mathrm{GL}_{n}(\mathbb{R})\to\mathcal{M}_{n}(\mathbb{R})$ with $f(B)=B^{-1}$ for every $B\in\mathrm{GL}_{n}(\mathbb{R})$, and it is the only one, by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol §equality. By Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets §restriction and Metric Space §metric-space, $(\mathrm{GL}_{n}(\mathbb{R}),d_{\mathrm{GL}})$ is a metric space, and $d_{\mathrm{GL}}(C,B)=d_{n}(C,B)$ for $C,B\in\mathrm{GL}_{n}(\mathbb{R})$. Let $C\in\mathrm{GL}_{n}(\mathbb{R})$ and $\varepsilon>0$. By the clause continuous with $C$ in place of $A$, there is $\delta>0$ such that every $B\in\mathcal{M}_{n}(\mathbb{R})$ with $d_{n}(C,B)<\delta$ satisfies $d_{n}(C^{-1},B^{-1})<\varepsilon$. So every $B\in\mathrm{GL}_{n}(\mathbb{R})$ with $d_{\mathrm{GL}}(C,B)<\delta$ satisfies

$$d_{n}(f(B),f(C))=d_{n}(B^{-1},C^{-1})=d_{n}(C^{-1},B^{-1})<\varepsilon,$$

by The Real Matrices with the Operator-Norm Distance Form a Complete Metric Space with Entrywise Convergence §metric (Metric Space §symmetric). Hence, by Continuous Map Between Metric Spaces §continuous-at with the subset taken to be the whole space $\mathrm{GL}_{n}(\mathbb{R})$, $f$ is continuous at $C$; as $C\in\mathrm{GL}_{n}(\mathbb{R})$ was arbitrary, $f$ is continuous, by Continuous Map Between Metric Spaces §continuous.