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!

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.   
