If $A$ is a continuous linear operator on $l^2$ such that $||A − T|| < 1$, show that $A$ is not invertible. Define $T : l^2 → l^2$ as
$T(x) = (0, x_1, x_2, · · ·)$ where $x=(x_1, x_2, · · ·) \in l^2$
If $A$ is a continuous linear operator on $l^2$ such that $||A − T|| < 1$, show that $A$ is not invertible.
Now I know that $T$ is not invertible again as $||A − T|| < 1$ hence $Id-A+T$ is invertible. Now if $A$ is invertible, does it imply that $T$ is invertible. Help how to solve this problem? 

Assume by contradiction that $A$ is invertible. Then, there exists an $x=(x_1,..,x_n,..)$ such that 
$$A(x)=(1,0,0,..0,...)$$
Then 
$$(A-T)(x)=(1,-x_1,-x_2,....,-x_n,..)$$
Therefore, since $\|A-T\| <1$, we have
$$\| (1,-x_1,..,-x_n,..) \| =\| (A-T)(x) \| < \| x \|=\| (x_1,..,x_n,..)\|$$
which is a contradiction.
