$\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$ for norm $\lVert•\rVert$ and nonsingular (matrix) $A$ and singular $B$? I have a question. For any norm $\lVert•\rVert$ and any nonsingular matrix $A$ and singular matrix $B$,
$$\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$$
How can we show that?
 A: It is not true for any square matrix norm. For example, if $\|\cdot\|_1$ is a norm, so is $\|\cdot\|_a=a\|\cdot\|_1$, for every $a>0$, and observe that for sufficiently small $a>0$,
we have $\|A^{-1}\|_a\|A-B\|_a=a^2\|A^{-1}\|_1\|A-B\|_1<1$.
However, it is true for every induced norm, i.e., norm of the form
$$
\|E\|=\sup_{\|x\|=1}\|Ex\|,
$$
where $\|\cdot\|$ is an arbitrary norm in $\mathbb R^n$.
Proof. Let $\|\cdot\|$ be an arbitrary norm in $\mathbb R^n$. Since $B$ is singular, there exists an $x_0\ne 0$, such that $Bx_0=0$. In fact, we may choose $x_0$, so that $\|x_0\|=1$.
So
$$
\|I-A^{-1}B\|=\sup_{\|x\|=1}\|(I-A^{-1}B)x\| \ge \|(I-A^{-1}B)x_0\|=\|x_0\|=1.
$$
But a property of induced norms is $\|EF\|\le \|E\|\|F\|$. Hence
$$
\|A^{-1}\|\|A-B\|\ge\|A^{-1}(A-B)\|=\|I-A^{-1}B\|\ge 1
$$
and thus
$$
\|A^{-1}\|\ge \frac{1}{\|A-B\|}.
$$
A: Assume $\|A^{-1}\| < \frac1{\|A-B\|}$. Then using the submultiplicativity of the norm we have
$$\|I-A^{-1}B\| = \|A^{-1}(A-B)\| \le \|A^{-1}\|\|A-B\| < 1$$
so $A^{-1}B$ is invertible with inverse $\sum_{n=0}^\infty (I-A^{-1}B)^n$ which converges since it converges absolutely. Hence $B$ is invertible as well, which is a contradiction.
