If $\|I-A\| \ge1$ does it follow that $A$ is singular? If $\|I-A\| \ge 1$ does it follow that $A$ is singular? I have proven the converse using the series $I + A + A^2 + \cdots$ but I am not sure about the other way around.
 A: Hint: Take $A=3\operatorname{Id}$.
A: No. Let $A=5I$. Then $$\|I-A\|=\|-4A\|=4$$ But $$A^{-1}=\frac{1}{5}I$$
A: We know that if
$\Vert I - A \Vert < 1, \tag 1$
then $A$ is nonsingular, since we have
$A^{-1} = (I - (I - A))^{-1} = \sum_0^\infty (I - A)^n, \tag 2$
and this series converges by virtue of (1).  For any $0 < r \in \Bbb R$ we have
$\vert 1 - r\Vert A \Vert \vert = \vert \Vert I \Vert - \Vert rA \Vert \vert \le \Vert I - rA \Vert, \tag 3$
which implies that $\Vert I - rA \Vert$ may be made arbitrarily large by sufficiently increasing $r$; but if $A$ is nonsingular, so is $rA$ for all non-zero $r$.  Furthermore, if we take
$r > 2 \Vert A \Vert^{-1}, \tag 4$
then $r \Vert A \Vert > 2$, so
$1 < r \Vert A \Vert - 1 = \vert r \Vert A \Vert - 1 \vert = \vert 1 - r \Vert A \Vert \vert \le \Vert I - rA \Vert, \tag 5$
thus for nonsingular $A$ we have in the family of operators
$\{rA \mid r > 2 \Vert A \Vert^{-1} \} \tag 6$
a class of examples with $rA$ nonsingular yet $\Vert I - rA \Vert \ge 1$.  And since this holds for any nonsingular $A$ . . . you get the idea.
In fact, since $rA$ is (non-)singular if and only if $A$ is the same, we may also take $A$ singular and obtain a family of singular operators satisfying (5).
