Question: Let $X $ be a Banach space and let $T $ and ${T^{ - 1}}$ belong to ${\cal B}\left( X \right)$. Prove that if $S \in {\cal B}\left( X \right)$ and $\left\| {S - T} \right\| < \frac{1}{{\left\| {{T^{ - 1}}} \right\|}}$, then $S^{ -1 } $ exists. (FOUNDATIONS OF MODERN ANALYSIS, Avner Friedman)
Here is a proof for matrix version. As we want to prove that $S$ is invertible, we are equivalently trying to prove $N\left( S \right) = \left\{ 0 \right\}$. The conditons we have can induce that
\[\begin{array}{l}
{\rm{ }}\left\| {S - T} \right\| < \frac{1}{{\left\| {{T^{ - 1}}} \right\|}}\\
\Rightarrow \left\| {{T^{ - 1}}} \right\|\left\| {S - T} \right\| < 1
\Rightarrow \left\| {{T^{ - 1}}\left( {S - T} \right)} \right\| < 1\\
\Rightarrow \left\| {{T^{ - 1}}S - I} \right\| < 1\\
\Rightarrow 1 - \left\| {{T^{ - 1}}S} \right\| < 1
\Rightarrow 0 < \left\| {{T^{ - 1}}S} \right\|
\end{array}\]
Actually, this may not help too much. We shall prove that $\forall x \ne 0,Sx \ne {\rm{0}}$, or equivalently, $\forall \left\| x \right\| \ne {\rm{0}},Sx \ne {\rm{0}}$. We then next try to do this:
\[\begin{array}{l}
\left\| x \right\| = \left\| {\left( {I - {T^{ - 1}}S + {T^{ - 1}}S} \right)x} \right\|
= \left\| {\left( {I - {T^{ - 1}}S} \right)x + \left( {{T^{ - 1}}S} \right)x} \right\|\\
\le \left\| {\left( {I - {T^{ - 1}}S} \right)x} \right\| + \left\| {\left( {{T^{ - 1}}S} \right)x} \right\|
\le \left\| {I - {T^{ - 1}}S} \right\|\left\| x \right\| + \left\| {\left( {{T^{ - 1}}S} \right)x} \right\|
\end{array}\]
Obviously, we can conclude that $0 < \left\| {\left( {{T^{ - 1}}S} \right)x} \right\|$ which is exactly what we want (since $\left\| {{T^{ - 1}}} \right\| > 0 $, $0 < \left\| {\left( {{T^{ - 1}}S} \right)x} \right\| < \left\| {{T^{ - 1}}} \right\|\left\| {Sx} \right\| \Rightarrow \left\| {Sx} \right\| > 0 \Rightarrow Sx \ne 0$).
Note: This theorem tells us that the lower bound (actually the greatest low bound, D.Lee & P.Y.Wu) of the distance between $T$ and the nearest sigular matrices. Once a matrix with the distance $\left\| {T - S} \right\|$ strictly lower than $\frac{1}{{\left\| {{T^{ - 1}}} \right\|}}$, it's then invertible.
Example: ${\left( {I - A} \right)^{ - 1}} = \sum\limits_{i = 0}^\infty {{A^i}} $ if $\left\| A \right\| < 1$. The condition is just to make sure that whether ${I - A}$ is invertible or not, comparing to the identity matrix $I$. That is, by the theorem above, if $\left\| {\left( {I - A} \right) - I} \right\| < \frac{1}{{\left\| {{I^{ - 1}}} \right\|}} = 1$ or $\left\| A \right\| < 1$, then the matrix ${I - A}$ is invertible and the expansion is meaningful.