Proving a matrix equality I have 2 matrices: $A \in R^{nxn}$ is a non-singular matrix and $B \in R^{nxn}$ is a singular matrix.  Here is the expression I need to prove:
$$||A - B|| \ge ||A^{-1}||^{-1}$$
I dont understand why it makes a difference to say B is singular.  Is there something special about the norm of a singular matrix?  I know the condition number of B = inf and its determinate is 0, thus has no inverse, but I dont know how these things can help me determine the inequality above.  I feel like I am missing a key point about singular matrices or something.
Also, would it be true to say in this instance that:
$$||A - B|| \ge ||A|| - ||B||$$
or should that be less than or equal to instead of greater than or equal to.
 A: The first inequality is actually a direct consequence of the following lemma:

Lemma. If $\|\cdot\|$ is a matrix norm and $\|I-X\|<1$, then $X$ is invertible.

I am not going to prove the lemma here. I think it can be found in many textbooks. Essentially, by using the properties of a matrix norm, one can first show that the partial sums of $Y=\sum_{k=0}^\infty (I-X)^k$ form a Cauchy sequence. Hence $Y$ converges. Then, one can show that $XY=I$ and in turn $X$ is invertible.
Since $A^{-1}B$ is singular, by the previous lemma, we have $1\le\|I-A^{-1}B\|=\|A^{-1}(A-B)\|\le\|A^{-1}\|\|A-B\|$. Thus we get your first inequality. Note that we have used the fact that $A^{-1}B$ is singular. Should $B$ be invertible, so be $A^{-1}B$ and we cannot make use of the contraposition of the previous lemma.
Your second inequality is correct. It is just the triangle inequality in disguise: $\|A\| = \|(A-B)+B\| \le \|A-B\|+\|B\|$. Yet we can refine it by a little: by interchanging the roles of $A$ and $B$, we have also $\|B-A\|\ge\|B\|-\|A\|$. So, putting the two inequalities together, we get $\|A-B\|\ge\left\lvert\left(\|A\|-\|B\|\right)\right\rvert$.
A: Some constraint on $B$ is needed, or we could take $B=A$ and deduce that
$$
 0 = \|A-A\| \ge \| A^{-1}\|^{-1}.
$$
Your inequality $\|A-B\| \ge \|A\| - \|B\|$ is a rearrangement of the triangle inequality and so holds for any two matrices.
And you should state what norm you are using. If we use the Frobenius norm and take $A$ to be the $3\times3$ identity matrix and let $B$ be the matrix we get
by setting $A_{3,3}=0$, then $\|A-B\|=1$ and $\|A^{-1}\|=3$. (In fact it is not clear to me that the inequality you want is true.)
A: The point is that if $A$ is nonsingular and $\|B - A\| < \|A^{-1}\|^{-1}$, then 
$B$ is also nonsingular.  In fact, if we write $B = A - T = A (I - A^{-1} T)$, then 
$$B^{-1} = \sum_{j=0}^\infty (A^{-1} T)^j A^{-1}$$
where the sum converges since $\|A^{-1} T\| \le \|A^{-1}\| \|T\| < 1$.
