Why does $\frac{1}{{\left\| {\left| {{A^{ - 1}}} \right|} \right\|}} \le \left\| {\left| B \right|} \right\|$? Let $A,B \in {M_n}$ suppose that the following statements are true:


*

*$A$ is nonsingular,

*$A+B$ is singular,

*$\left\| {\left| . \right|} \right\|$ is matrix norm.


Why is it true that: $\frac{1}{{\left\| {\left| {{A^{ - 1}}} \right|} \right\|}} \le \left\| {\left| B \right|} \right\|$?
 A: If $A$ is nonsingular and $A+B$ singular then $I+A^{-1}B$ is singular. This means that $I+A^{-1}B$ has at least one zero eigenvalue and therefore there exists some $0\neq v\in\mathbb{R}^n$ such that 
$$(I+A^{-1}B)v=0$$ which yields
$$\frac{\|A^{-1}Bv\|}{\|v\|}=1$$
The above identity ensures by induced matrix norm definition (the supremum of this ratio over all nonzero vectors $v$) that
$$\|A^{-1}B\|\geq 1$$
and by norm properties
$$\|A^{-1}\|\|B\|\geq\|A^{-1}B\|\geq 1$$
A: Note that $I + A^{-1}B$ is singular.  We therefore note that
$$
\left\| {\left| A^{-1}B \right|} \right\| \geq 1
$$
(How?) Thus, we have
$$
1 \leq \left\| {\left| A^{-1} B \right|} \right\| \leq 
\left\| {\left| A^{-1} \right|} \right\| \left\| {\left| B \right|} \right\|
$$
The conclusion follows.

Note: This also works if $|\|\cdot\||$ is any submultiplicative matrix norm.  In particular, we note that $|\|A\|| \geq \rho(A)$ for any matrix $A$ since for any eigenvector $x$ associated with eigenvalue $\lambda$, we have
$$
|\lambda| \cdot |\|\pmatrix {x&\cdots & x}\|| =
|\|A \pmatrix {x&\cdots & x}\|| \leq 
|\|A\|| \cdot |\|\pmatrix {x&\cdots & x}\||
$$
A: Let $w \neq 0$ be such that $(A+B)w = 0$. Then,
$$\left\| {\left| A^{-1}\right|} \right\| = \max_{v\neq 0} \frac{|A^{-1}v|}{|v|} \geq \frac{|A^{-1}Aw|}{|Aw|} = \frac{|w|}{|Aw|} = \left(\frac{|Bw|}{|w|}\right)^{-1} \geq \left(\max_{v\neq 0} \frac{|Bv|}{|v|}\right)^{-1} = \left\| {\left| B \right|} \right\|^{-1}$$
