Let $V$ be a normed space and let $\alpha ∈ End(V )$ have an induced norm satisfying $\|\alpha\| < 1$. Show that $\sigma_1 -\alpha \in Aut(V )$.
Let $\|\:.\|$ be a norm on $V$ and $\alpha \in End(V)$. Assume that $\|\:.\|_*$ is the induced norm defined by $\|\:.\|$ and satistfying the condition $\|\alpha\|_* < 1$. Hence,
$$\|\alpha\|_*=\max\{\frac{\|\alpha(v)\|}{\|v\|}\big{|}0\neq v\in V\}<1,$$ which implies that $\|\alpha(v)\|<\|v\|$ for any nonzero vector $v\in V$. Let $0\neq v\in V$, then since $$\|v\|=\|v-\alpha(v)+\alpha(v)\|\leq\|v-\alpha(v)\|+\|\alpha(v)\|$$ $$<\|(\sigma_1-\alpha)(v)\|+\|v\|,$$
and so $\|(\sigma_1-\alpha)(v)\|>0$, which implies that $v\notin ker(\sigma_1-\alpha )$.therefore, $(\sigma_1-\alpha)$ is monic.
If $V$ was finitely generated, I could easily say that $\sigma_1-\alpha\in Aut(V)$, but I don't know what to say when $V$ is an arbitrary normed space.
This is not true without assuming that $V$ is complete. Counterexample: let $V$ be the space of sequences that are eventually $0$ (meaning the sequence has finitely many nonzero elements), equipped with the $\ell^2$ norm $\sqrt{\sum_{n=1}^\infty x_n^2}$ (or any other $\ell^p$ norm). Let $\alpha((x_1,x_2,\dots)) = (0, x_1/2, x_2/2, \dots)$ be the forward shift times $1/2$. Clearly $\|\alpha\|_*=1/2$. But the map $\sigma_1-\alpha$ is not surjective; the basis vector $e_1=(1,0,0,\dots)$ is not in its image. Indeed, if $(\sigma_1-\alpha)(x) = e_1$, then $x_1=1$, $x_2=1/2$, $x_3=1/4$, and so on; but there is no such element in $V$ (where the sequences are eventually zero).
If $V$ is complete, then the claim follows from Banach's fixed point theorem. Indeed, for any given $y\in V$ the map $x\mapsto y+\alpha(x)$ is appropriately contracting, hence has a fixed point. This fixed point satisfies $x-\alpha(x) = y$.