I am working with the following lemma given in the book Topics in Banach Spaces Theory:
Let ${(x_n)}_{n=1}^{\infty}$ be a basic sequence in Banach Space $X$. Suppose that there exists a linear functinal $x^{\ast} \in X^{\ast} $ such that $x^{\ast}(x_n)=1$ for all $n\in \mathbb{N}$. If $u\notin [x_n]$ then the sequence ${(x_n +u)}_{n=1}^{\infty }$ is basic.
This is the proof:
Since $u\notin [x_n]$, without loss of generality we can assume $x^{\ast}(u)=0$. Let $T:X\to X$ given by $\:T(x)=x^{\ast}(x)u$. Then $I_X+T$ is invertible with inverse $I_X-T$. Since $(I_X+T)(x_n)=x_n+u$, the sequences $\{x_n\}_{n=1}^{\infty}$ and $\{x_n+u\}_{n=1}^{\infty}$ are congruent, hence $\{x_n+u\}_{n=1}^{\infty}$ is basic.
I don't understand why $I_x+T$ is invertible. According to the Theorem about the existence of the inverse operator (which can be found on the page 329 of this book) if $\lVert I_X-(I_X+T)\lVert = \lVert T\lVert<1$ then we are guranteed the existce of the inverse and $${(I_x+T)}^{-1}=\lim_{n \to \infty} [I_X+ (-T)+ {(-T)}^2+\dots+ {(-T)}^n ]$$ Based on our assumption that $x^{\ast}(u)=0$ then $$T^2(x)=T(T(x))=T(x^{\ast}(x)u)=x^{\ast}(x^{\ast}(x)u)u=x^{\ast}(x)x^{\ast}(u)u=0$$ and thus ${(-T)}^2$=0. Hence we would have $T^3=T(T^2)=0 \quad \text{and} \quad \dots T^n=0$ and therefore $${(I_x+T)}^{-1}=\lim_{n \to \infty} [I_X+ (-T)]=I_X-T$$ If I knew that $\lVert T\lVert<1$ then everything is ok, but why is it so?
The operator $I+T$ is a rank-one perturbation of identity, displacement along $u$. Importantly, the amount of displacement is constant on every line parallel to $u$; this allows us to return to original position by applying the opposite displacement. Here's this statement in greater generality, for possible nonlinear operators (in a space over field $\mathbb K$).
If $f:X\to \mathbb K$ is such that $f(x+tu)=f(x)$ for all $x\in X$ and $t\in\mathbb K$, then the maps $x\mapsto x\pm f(x)u$ are inverses of each other.
Indeed, $$(x+f(x)u)-f(x+f(x) u )u = (x+f(x)u)-f(x)u =x$$