Upper bound of a norm of an inverse of a matrix I'm looking for a way to bound the norm of an inverse of a matrix (specifically considering infinity-norm). I know that the norm of an inverse of a matrix has the relationship,
$||A^{-1}|| \geq \left(||A|| \right)^{-1}$ (Eq. 1)
but this provides a lower bound. Are there any useful equations or identities that could form the upper bound on this norm instead? (For example, in which situations does the equality for Eq. 1 hold?)
I do not have knowledge of eigenvalues/singular values of the matrix (that's why I'm using the infinity-norm instead). The matrix is complex and has the form $A := \hat{I} - {\mathrm i} \alpha B$, where $\hat{I}$ is the identity, ${\mathrm i}$ is the imaginary unit, $\alpha$ is a real constant, and $B$ is a matrix with real elements and a symmetric structure.
Attempt:
I know we can split up the norm of a sum of a matrix into the sum of norms of matrices, i.e.
$||A+B|| \leq ||A|| + ||B||$. (Eq. 2)
While this provides an upper bound on $(||A||)^{-1}$ if we use Eq. 1 to decompose the norm of an inverse matrix, I don't know how it relates to the upper bound on $||A^{-1}||$.
 A: My answer does not address this question in general, but it still might be adequate.
Let $\Delta=B-A$, where $A$, $B$ and $\Delta$ are complex square matrices. In this scenario I want to compute a bound on $\|A^{-1}-B^{-1}\|$ as a function of $\|\Delta\|=\|A-B\|$.
Since $A^{-1}\Delta B^{-1}=A^{-1}\left(B-A\right) B^{-1}=A^{-1} - B^{-1}$, it follows that
\begin{align}
\| A^{-1} - B^{-1} \| &= \|A^{-1}\Delta B^{-1}\| \\
&\leq \|A^{-1}\Delta\| \| B^{-1}\|  \ , \tag 1 \label{eq:1}
\end{align}
due to the fact that matrix norms are submultiplicative. Furthermore, since $B^{-1}=A^{-1}-A^{-1}\Delta B^{-1}$ we can use the triangle inequality and submultiplicative property (once again) to obtain
\begin{align}
\|B^{-1}\| &\leq \|A^{-1}\|+\|A^{-1}\Delta B^{-1}\| \\
&\leq \|A^{-1}\|+\|A^{-1}\Delta\| \|B^{-1}\| \ ,
\end{align}
which is equivalent to
\begin{align}
\|B^{-1}\| &= \|\left(A+\Delta \right)^{-1}\| \\
&\leq \frac{\|A^{-1}\|}{1-\|A^{-1}\Delta\|} \ . \tag 2 \label{eq:2}
\end{align}
Finally, if we plug \eqref{eq:2} back in \eqref{eq:1} we get the desired upper bound
\begin{align}
\|A^{-1}-B^{-1}\|&\leq\frac{\|A^{-1}\|\|A^{-1}\Delta\|}{1-\|\|A^{-1}\Delta\|} \\
&\leq \frac{\|A^{-1}\|\|A^{-1}\|\|\Delta\|}{1-\|A^{-1}\Delta\|} \\
&\leq \frac{\|A^{-1}\|^2\|A-B\|}{1-\|A^{-1}\Delta\|} \ .
\end{align}
Full disclosure: This answer is based on the beginning of section "5.8 Condition numbers: inverses and linear systems" from "Matrix Analysis" by Roger A. Horn and Charles R. Johnson (2nd Edition, p. 381).
