Definition 1.5.1 (Open ball). For any $x\in \R^n$ and any $r>0$, the open ball of radius $r$ around $x$ is the subset

$$ B_r(x) = \{y\in \R^n \text{ such that } |x-y|< r\}. $$

Note that $|x-y|$ must be less than $r$ for the ball to be open; it cannot be = $r$. In dimension 1, a ball is an interval.

Definition 1.5.2 (Open set of $\R^n$). ****A subset $U\subset \R^n$ is open if for every point $x\in U$, there exists $r>0$ such that the open ball $B_r(x)$ is contained in $U$.

Geometric Series of Matrices

Let $A$ be a square matrix. If $|A| < 1$, the series $S = I+A+A^2+\dots$ converges to $(I-A)^{-1}$.

Proof.

$$ S_k = I+A+A^2+\dots+A^k\\ S_kA = A+A^2+\dots+A^k+A^{k+1} $$

By substracting $S_kA$ from $S_k$, we get that $S_k-S_kA =S_k(I-A)= I-A^{k+1}$. By the Cauchy-Schwarz inequality, we know that $|A^{k+1}| \leq |A|^k|A|=|A|^{k+1}$. This tells us that the series of numbers $\sum_{k=1}^{\infty} |A|^k$ converges when $|A| < 1$, hence, $S$ converges. Thus,

$$ S(I-A)=\lim_{k\to \infty}S_k(I-A) = \lim_{k\to\infty}(I-A^{k+1}) $$

$$ =I-\lim_{k\to\infty}A^{k+1} = I $$

This equation shows that the inverse of $I-A$ is $S$.

We now have to prove that the set of invertible $n\times n$ matrices is open.

Proof. Suppose $B$ is invertible, and $|H| < 1/|B^{-1}|$. Then, $|-B^{-1}H| < 1$, so $I+B^{-1}H$ is invertible, and

$$ (I+B^{-1}H)^{-1}B^{-1} = (B(I+B^{-1}H))^{-1}=(B+H)^{-1}. $$

Thus if $|H| < 1/|B^{-1}|$, the matrix $B+H$ is invertible, giving an explicit neighborhood of $B$ made up of invertible matrices.