The question is that:
We have two matrices $A,B\in \mathbf{C}^{n\times n}$, A is nonsingular and B is singular, let $||\cdot ||$ be $\textbf{any}$ matrix norm, prove
$||A-B||\geq 1/||A^{-1}||$.
My idea is
$||(A-B)||\cdot||A^{-1}||\geq||A^{-1}(A-B)||=||I-A^{-1}B||$
And I am confused here. How to calculate the right side?
$\color{red}{Note:}$ the original question may be wrong when the norm doesn't have sub-multiplicative property. See the comments below.
Thanks for everybody's help!
Best Answer
HINT:
$B= A+ (B-A) = A\cdot( I - A^{-1}(A-B) ) = A \cdot (I - C)$
where $C = A^{-1}(A-B)$. Assume that $||A-B||< ||A^{-1}||^{-1}$. Then $||C||\le ||A^{-1} || \cdot ||A-B|| < 1$. Now use the following fact:
If $C$ is a matrix with norm $||C||<1$ then the series $\sum_{n\ge 0} C^n$ is absolutely convergent and its sum is the inverse of $I-C$. In particular: $||C||<1$ implies $(I-C)$ invertible.
Use the above to conclude $B$ is invertible, contradiction.