Showing that $|||A-B|||\geq \frac{1}{|||A^{-1}|||}$? $A,B\in M_n$, $A$ is non-singular and $B$ is singular. $|||\cdot|||$ is any matrix norm on $M_n$, how to show that $|||A-B||| \geq \frac{1}{|||A^{-1}|||}$?
The hint is let $B=A[I-A^{-1}(A-B)]$, but I don't know how to use it.
Appreciate any help!
update: is $\geq$,not $\leq$.Sorry!
 A: Hint for operator norm: notice that $\frac{1}{|||A^{-1}|||}=\inf_{v\neq 0} \lVert Av\rVert/\lVert v\rVert$, and then consider $(A-B)v$ for $v\in \ker B$.
A: Let's sharpen the hint to $A^{-1}B = I - A^{-1}(A-B)$. First you should check that this identity is correct.
Now pick any $v$ such that $Bv = 0$ and $\|v \| = 1$. By assumption, such a $v$ exists. Apply both sides of the identity, play around with it, take norms, see if you can get something that resembles the statement that you want to prove.
Remember the definition of a matrix norm: $|||C||| = \sup_{\|x\| = 1} \|Cx\|$. There is also a formula that relates $|||CD|||$ to $|||C|||$ and $|||D|||$. Check your notes and try to use it.   
A: This is wrong. In what follows, we assume that we are working with the operator norm. 
If $A$ is non-singular, it is easy to check that $P$ is non-singular for all $P\in B\left(A;\frac{1}{\Vert A^{-1}\Vert}\right)$.
The proof runs as follows. Let $P\in B\left(A,\frac{1}{\|A^{-1}\|}\right)$. We
shall prove that $P\in GL_n(\mathbb{R})$. For each $x\in\mathbb{R}^n$,
\begin{eqnarray*}
\|x\|&=&\|A^{-1}(Ax)\|\leqslant \|A^{-1}\|\|Ax\|\\& \leqslant &
\|A^{-1}\|\left(\|(A-P)x\|+\|Px\|\right)\leqslant
\|A^{-1}\|\left(\|A-P\|\|x\|+\|Px\|\right).
\end{eqnarray*}
Thus,
$$
\|x\|\leqslant \frac{\|A^{-1}\|}{1-\|A-P\|\|A^{-1}\|}\|Px\|,\text{
for all }x\in\mathbb{R}^n.
$$
This implies that $P\in GL_n(\mathbb{R})$ because if $ Px=0$, we have
$x=0$. Thus we have proved that $
B\left(A,\frac{1}{\|A^{-1}\|}\right)\subset GL_n(\mathbb{R})$.
