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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?

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    In fact, A can neither be injective nor surjective. @N.S. shows the surjectivity part and for the injectivity we can proceed just the same as in this answer but replace $x$ with a nonzero kernel element of $A$. – 2017-01-08

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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.