Proof of Continuity of the Inverse of a Continuous Matrix Function

For a real $k\times k$ matrix $U$ write $|U|_{e}=\max_{i,j}|U_{ij}|$; by claim 2 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, $|UV|_{e}\le k\,|U|_{e}\,|V|_{e}$. Products are rearranged with Associativity of the Matrix Product, inverses are unique by Uniqueness of the Matrix Inverse, and transpose identities are claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals. We also use limit arithmetic for real sequences: if $u_n\to u$ and $v_n\to v$ then $u_n+v_n\to u+v$ and $u_nv_n\to uv$, by the standard estimates $|u_nv_n-uv|\le|u_n|\,|v_n-v|+|v|\,|u_n-u|$ and $|(u_n+v_n)-(u+v)|\le|u_n-u|+|v_n-v|$ (a convergent sequence being bounded).

Claim 1. Fix $t\in[a,b]$ and put $\beta=|M(t)^{-1}|_{e}$. For $s\in[a,b]$ define

By continuity of the entries of $M$, given any $\eta\in(0,\tfrac12]$ there is $\delta>0$ such that $|s-t|<\delta$ (with $s\in[a,b]$) implies $q(s)\le\eta\le\tfrac12$. Fix such an $s$ and write $E=E(s)$, $q=q(s)$.

Geometric series. Let $S_N=\sum_{n=0}^{N}(-E)^{n}$ (matrix powers; $(-E)^{0}=I_k$, the identity matrix). By induction from the product bound, $|(-E)^{n}|_{e}\le k^{\,n-1}|E|_{e}^{\,n}=q^{n}/k$ for $n\ge1$. Hence for $N<N'$, $|S_{N'}-S_N|_{e}\le\sum_{n=N+1}^{N'}q^{n}/k\le q^{N+1}/\bigl(k(1-q)\bigr)$, so each entry of $(S_N)_N$ is a Cauchy sequence (the geometric tail bound, with $q\le\tfrac12$) and converges by Every Cauchy Sequence of Real Numbers Converges; let $S$ be the entrywise limit, so that also $|S-I_k|_{e}\le\sum_{n\ge1}q^{n}/k\le2q/k$ (limits preserve the partial-sum bounds).

Telescoping, $(I_k+E)S_N=S_N+ES_N=\sum_{n=0}^{N}(-E)^{n}-\sum_{n=0}^{N}(-E)^{n+1}=I_k-(-E)^{N+1}$, whose entries tend to those of $I_k$ since $|(-E)^{N+1}|_{e}\le q^{N+1}/k\to0$. Each entry of $(I_k+E)S_N$ is a finite sum of products of entries, so by limit arithmetic it converges to the corresponding entry of $(I_k+E)S$; hence $(I_k+E)S=I_k$, and symmetrically $S(I_k+E)=I_k$ (the same telescoping on the other side). Thus $I_k+E$ is invertible with inverse $S$.

Identification. $M(t)(I_k+E)=M(t)+\bigl(M(s)-M(t)\bigr)=M(s)$. Hence $S\,M(t)^{-1}$ is a two-sided inverse of $M(s)$: $\bigl(SM(t)^{-1}\bigr)M(s)=S\,M(t)^{-1}M(t)(I_k+E)=S(I_k+E)=I_k$ and $M(s)\bigl(SM(t)^{-1}\bigr)=M(t)(I_k+E)S\,M(t)^{-1}=M(t)M(t)^{-1}=I_k$. By uniqueness, $M(s)^{-1}=S\,M(t)^{-1}$, and

Given $\varepsilon>0$, choose $\eta=\min(\tfrac12,\varepsilon/(2\beta+1))$ and the corresponding $\delta$: then $|s-t|<\delta$ implies every entry of $M(s)^{-1}$ is within $\varepsilon$ of the corresponding entry of $M(t)^{-1}$. Since $t$ was arbitrary, the entries of $t\mapsto M(t)^{-1}$ are continuous on $[a,b]$.

Claim 2. Let $M=M(t)$ be symmetric positive definite (positive definiteness alone already forces invertibility: if $Mx=0$ with $x\ne0$ then $x\cdot(Mx)=0$, impossible; so the hypothesis of claim 1 is not enlarged). By claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, $(M^{-1})^{\top}M^{\top}=(M M^{-1})^{\top}=I_k^{\top}=I_k$ and $M^{\top}(M^{-1})^{\top}=(M^{-1}M)^{\top}=I_k$; with $M^{\top}=M$ and uniqueness of the inverse (Uniqueness of the Matrix Inverse), $(M^{-1})^{\top}=M^{-1}$: the inverse is symmetric. Positive definiteness: let $x\ne0$ and put $y=M^{-1}x$; then $y\ne0$ (otherwise $x=My=0$), and by the transpose-dot identity and symmetry,