What does this theorem mean?

Question

Let $(V,\|\cdot\|)$ be a finite-dimensional normed space.

Define $\|T\|_\mathrm{op}=\sup\{\|T(x)\|:\|x\|≦1\}$, for all linear operators on $V$

Define $\Omega$ to be the set of all invertible linear operators on $V$.

With these notations above, below are theorems i'm curious to know what they do mean:

Theorem1.

Let $A\in\Omega$ and $B$ be a linear operator on $V$.

If $\|B-A\|_\mathrm{op}\|A^{-1}\|_\mathrm{op}<1$, then $B\in\Omega$.

Theorem2.

$F:\Omega\rightarrow\Omega:T\mapsto T^{-1}$ is continuous.

What do they actually mean? I have no idea why these are useful theorems and how i could visualize these theorems. Please help. Thank you in advance :)

Answer 1

Visualizing such things might prove hard, since we're dealing with vectors in $n^2$ dimensions for some $n$. The theorems are important since they provide, along with the contraction principle, a very good proof of the inverse functions theorem which Rudin gives in full detail.

To prove the first theorem, we start with $\lVert A-B\rVert \rVert A^{-1}\rVert<1$. I claim this implies $B$ is injective, so that it is invertible. Indeed, we know that for any matrix, $$\lVert Bx\rVert \leqslant \lVert B\rVert \lVert x\rVert$$

I say the inequality implies $\lVert Bx\rVert\geqslant C\lVert x\rVert$, for some $C>0$, so that $Bx=0\implies x=0$ and $B$ is injective. Indeed, by submultiplicativity and the triangle inequality $$\lVert Ax\rVert \leqslant \lVert A-B\rVert \lVert x\rVert +\lVert Bx\rVert$$

But $\lVert A^{-1}\rVert^{-1}\lVert x\rVert=\lVert A^{-1}\rVert^{-1}\lVert A^{-1}Ax\rVert\leqslant \lVert Ax\rVert$ by submultiplicativity. This together gives $$(\lVert A^{-1}\rVert^{-1} -\lVert A-B\rVert)\lVert x\rVert \leqslant \lVert Bx\rVert$$

Since we made it so the term in parenthesis is positive, we're done. In general, one needs to observe there aren't many choices once we have a set up: certain terms must appear in inequalities, and there are just so many things we know about inequalities between norms that we should try using those. The second statement is easy once you see $A^{-1}$ is a rational function on the entries of $A$.

Answer 2

Please refer to the exercise 7 on page 155 of Folland's real analysis. You will find more useful information.

Let $X$ be a Banach space.

1. If $T\in L(X,X)$ and $\|I-T\|<1$ where $I$ is the identity operator, then $T$ is invertible; in fact, the series $\sum_0^\infty(I-T)^n$ converges to $(I-T)^{-1}$.
2. If $T\in L(X,X)$ is invertible and $\|S-T\|<\|T^{-1}\|^{-1}$, then $S$ is invertible. Thus the set of invertible operators is open in $L(X,X)$