Exercise :
Consider the infinite-dimensional system of equations : $$x_i - \sum_{j=1}^\infty a_{ij}x_j = b_i, \quad i=1,2,3,\dots$$ We suppose that $b=(b_1,b_2,\dots) \in \ell^\infty$ and that it exists $0<\theta<1$ such that : $$\sup_i \sum_{j=1}^\infty |a_{ij}|\leq \theta$$ Show that the system of equations given has a unique solution $x=(x_1,x_2,\dots) \in \ell^\infty$.
Attempt :
In previous exercises and lessons, I have proved the following lemma :
Let $X$ be a Banach space and $T \in B(X)$ with $\|T\| \leq \theta < 1$. Then, if $y \in X$, the equation $x = y + Tx$ has a unique solution $x \in X$.
My initial idea is to work over manipulating the exercise to reach the statements of the lemma above.
The equation for the infinite-dimensional system of equations given, can be rewritten as :
$$x_i - \sum_{j=1}^\infty a_{ij}x_j = b_i \Rightarrow x = b + Tx$$
where $x = (x_1,x_2,\dots)$ and $T$ be an operator, such that :
$$Tx = \sum_{j=1}^\infty a_{ij}x_j$$
Now, I observe that $\ell^\infty$ is a Banach space and since $b \in \ell^\infty$ then $x - Tx \in \ell^\infty$ which means that $x \in \ell^\infty$ and $T$ is an operator defined over $\ell^\infty$.
Now, I need to prove that $T \in B(\ell^\infty)$, which means that $T$ is a bounded linear operator $T : \ell^\infty \to \ell^\infty$ and that $\|T\| < 1$.
For the case of linearity, let $x,y \in \ell^\infty$ and $\lambda \in \mathbb R$. Then, it is
$$T(\lambda x + y) = \sum_{j=1}^\infty a_{ij}(\lambda x_j + y_j) = \sum_{j=1}^\infty \lambda a_{ij} x_j + \sum_{j=1}^\infty a_{ij}y_j$$
$$=$$
$$\lambda \sum_{j=1}^\infty a_{ij}x_j + \sum_{j=1}^\infty a_{ij}y_j = \lambda T x + Ty$$
which proves that $T$ is a linear operator.
Now, for the case of $T$ being bounded :
$$\|Tx\|_\infty = \bigg\|\sum_{j=1}^\infty a_{ij}x_j \bigg\|_\infty \leq \sum_{j=1}^\infty \|a_{ij}x_j\|_\infty \leq \sum_{j=1}^\infty |a_{ij}| \|x\|_\infty$$
But, it is
$$\sup_i \sum_{j=1}^\infty |a_{ij}|\leq \theta < 1$$
thus, by combining the two last results above, we get :
$$\|Tx\|_\infty \leq \theta \|x\|_\infty$$
which proves that $T$ is a bounded linear operator $T \in B(\ell^\infty)$ and thus the equation we transformed has a unique solution $x \in \ell^\infty$ which means that the initial infinite-dimensional system of equations has a unique solution $x=(x_1,x2,\dots) \in \ell^\infty$.
Question : Is my approach correct and rigorous enough ? Any recommendations on what I can improve or any mistakes ? I would appreciate any input or approvement of my approach.