1.5. Limits and Continuity

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.

Problem 1.5.23

A. Let $A$ be an $n \times n$ matrix. What does it mean to say that

$\lim_{B\to A}(A-B)^{-1}(A^2-B^2) =C$

exists?

It means that $\forall \varepsilon > 0$, $\exists \delta > 0$ such that $\forall x \in X$, $|x-x_0| < \delta \implies |f(x) -a| < \varepsilon$.

Let $X=A-B$.

$\lim_{\|X\| \to 0}X^{-1}(A^2-(A-X)^2)$ $=\lim_{\|X\| \to 0}X^{-1}(A^2-A^2+AX+XA-X^2)$ $=\lim_{\|X\|\to0}X^{-1}(AX+XA-X^2)$ $\lim_{\|X\|\to 0}+X^{-1}AX+X^{-1}XA-X^{-1}X^2$ $=\lim_{\|X\|\to 0}X^{-1}AX+A$

If $AX$ is communicative, then the limit is $2A$. Therefore, if $AB = BA$, then the limit exists.

B. Does the limit exist when $A = \begin{bmatrix}1&0\\0&1\end{bmatrix}$?

Yes. Because $IA = AI$, the limit exists and is $2I$.

C. Does the limit exist when $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$?

Yes. To illustrate, let $B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. $X= \begin{bmatrix}-a&1-b\\1-c&-d\end{bmatrix}$.