"Functional Analysis" by Kesavan, Exercise 2.32 (b)
Define $T, S\colon \ell_2 \to \ell_2$ by
\begin{align*}T(x) &= (0, x_1, x_2, \dots) \\ S(x) &= (x_2, x_3, \dots)\end{align*}
where $x = (x_1, x_2, \dots) \in \ell_2$.If $A$ is a continous linear operator on $\ell_2$ such that
$\Vert A – T\Vert < 1$, show that $A$ is also not invertible.
Deduce that $\mathcal{G}$, the set of invertible linear operators
in $\mathcal{L}(V)$, is not dense in $\mathcal{L}(V)$, if $V$ is
infinite dimensional.
Part (a) of the question was to show that $T, S$ are continuous linear operators, and that $ST = I$ while $TS \neq I$, which I could do. This shows that $T$ and $S$ are not invertible.
For the second statement, I have an intuition that if $\mathcal{G}$ were dense in $\mathcal{L}(V)$, then there would exist a sequence of invertible linear operators converging to $T$, which woud mean that some $A_n \in \mathcal{G}$ must satisfy $\Vert A_n – T\Vert < 1$, contradicting the first statement.
Best Answer
We have $$\|SA-I\|=\|S(A-T)\|\le \|A-T\|<1$$ Thus $SA$ is invertible. As $S$ is not invertible hence $A$ is not invertible as well. This shows that the distance of $T$ to the set of invertible operators is greater or equal $1,$ i.e. invertible operators are not dense.
Actually the distance of $ T$ to invertible operators is equal $1$ because the operator $U:=T-(1+n^{-1})I$ is invertible and $\|U-T\|=1+n^{-1}.$
Remark It is well known that $\sigma(T)=\{z\in \mathbb{C}\,:\,|z|\le 1\}$. Thus $0\in {\rm int}\,\sigma(T).$ Obviously if $z\in {\rm bd}\,\sigma(A)$ then $A-zI$ can be approximated by invertible operators. But the converse is not true in general. Namely any normal operator $A,$ such that $0\in {\rm int}\,\sigma(A),$ can be approximated by invertible operators.