TITLE: An old test question proving $\|\mathbf{B} - \mathbf{A}\| \lt \frac{1}{\|\mathbf{A}^{-1}\|}$ implies invertiblity of $\mathbf{B}$
QUESTION [6 upvotes]: I have an old test question that I am not sure about and would like some idea. It is from a Numerical Analysis class.
Suppose that $A$ is an invertible $n$-by-$n$ matrix. Prove that for every $n$-by-$n$ matrix $B$, the inequality
$$ \|\mathbf{B} - \mathbf{A}\| \lt \frac{1}{\|\mathbf{A}^{-1}\|}$$
implies that $B$ is invertible.
I answered the test question correctly, but am also interested in what this particular equation could be used for.
EDIT:
Here is what I did on my test (the test did give a hint to use the contrapositive):
Suppose $A$ is invertible and $B$ is singular. Then for some $x \ne 0$, we have $Bx = 0$
Note that
$$\begin{align}
0  \ne & \| x\| & = & \| A^{-1}Ax\| \\
&&  = & \|A^{-1}(Ax - Bx)\| \\
&&  = & \|A^{-1}(A-B)x\| \\
&&  \le & \|A^{-1}\|\cdot\|(A-B)x\| \\
\implies &  \|x\| & \le &  \|A^{-1}\|\cdot\|A-B\| \cdot \|x\|\\
\\
\implies & \frac{ \|x\|}{\|x\|} & \le &  \|A^{-1}\|\cdot\|A-B\| \cdot \frac{\|x\|}{\|x\|}  \,\,\,\,\text{since $\|x\| \ne 0$} \\
\implies & 1 & \le & \|A^{-1}\|\cdot\|A-B\| \\
\implies & \frac{ 1}{\|A^{-1}\|} & \le & \|A-B\| \\
\therefore & B \;\text{singular} \implies && \|A-B\| \ge \frac{1}{A^{-1}} \\
\text{By contrapositive argument,}&\text{ we have} \\
&\|A-B\|&  \lt &\frac{1}{\|A^{-1}\|} \implies \text{$B$ is invertible}\\
\end{align}
 $$

REPLY [2 votes]: This result is from Ortega and Rheinboldt, 1970, Iterative Solution of Nonlinear Equations in Several Variables (Lemma 2.3.2). There are sure to be other proofs, but I like this one.
We start with the following preliminary result (which you can find in the above book).
$\textbf{Lemma 1}$ Let $B \in L(\mathbb{R}^{n})$ and assume that $\rho(B) < 1$, where $\rho(B)$ is the spectral radius of $B$. Then $(I - B)^{-1}$ exists.
Using this we now use the fact that
$\displaystyle || A - B || < \frac{1}{|| A ||^{-1}}$
and note that
$|| I - A^{-1}B || = || A^{-1}(A - B) || \leq || A^{-1} ||\,|| A - B || < 1.$
Hence, using Lemma 1 we have that
$A^{-1}B = I - (I - A^{-1}B)$
is invertible, so therefore $B$ must also be invertible