An old test question proving $\|\mathbf{B} - \mathbf{A}\| \lt \frac{1}{\|\mathbf{A}^{-1}\|}$ implies invertiblity of $\mathbf{B}$ 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
$$\begin{align}
0  \ne & \| x\| & = & \| A^{-1}Ax\| \\
&&  = & \|A^{-1}(Ax - Bx)\| \\
&&  = & \|A^{-1}(A-B)x\| \\
&&  \le & \|A^{-1}\|\cdot\|(A-B)x\| \\
\implies &  \|x\| & \le &  \|A^{-1}\|\cdot\|A-B\| \cdot \|x\|\\
\\
\implies & \frac{ \|x\|}{\|x\|} & \le &  \|A^{-1}\|\cdot\|A-B\| \cdot \frac{\|x\|}{\|x\|}  \,\,\,\,\text{since $\|x\| \ne 0$} \\
\implies & 1 & \le & \|A^{-1}\|\cdot\|A-B\| \\
\implies & \frac{ 1}{\|A^{-1}\|} & \le & \|A-B\| \\
\therefore & B \;\text{singular} \implies && \|A-B\| \ge \frac{1}{A^{-1}} \\
\text{By contrapositive argument,}&\text{ we have} \\
&\|A-B\|&  \lt &\frac{1}{\|A^{-1}\|} \implies \text{$B$ is invertible}\\
\end{align}
 $$
 A: It shows that the set of invertible matrices is an open set: If a matrix $A$ is invertible, there exists a neighborhood such that all matrices $B$ in this neighborhood are invertible as well.
Furthermore this proof can fairly easily be generalized to the infinite-dimensional case, where $A$ and $B$ are bounded linear operators on a Banach space.
A: This is a standard result. If $\|\Delta\| <1$, and $\|\cdot\|$ is a sub-multiplicative norm, then we have
$\|I+\Delta+...+\Delta^n\| \le 1 +\|\Delta\| + ... +  \|\Delta\|^n = { 1-\|\Delta\|^{n+1} \over 1-\|\Delta\|}  \le { 1 \over 1-\|\Delta\|}$. Hence we have that $D = \sum_{k=0}^\infty \Delta^k$ exists, and furthermore, since $(I-\Delta) (I+\Delta+...+\Delta^n) = I-\Delta^{n+1}$, we see that $(I-\Delta) D = I$. That is, $D$ is the inverse of $I-\Delta$.
Now suppose $A$ is invertible, and $\|H\| < {1 \over \|A^{-1}\|}$. Then, since we have $A+H = A(I+A^{-1}H)$, we see that $A+H$ is invertible iff $(I+A^{-1}H)$ is invertible. Furthermore, since $\|A^{-1}H\| \le \|A^{-1}\| \|H\|$, we have $\|A^{-1}H\| < 1$. Hence $(I+A^{-1}H)$ is invertible.
In the question, let $H=B-A$.
A: 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
