Proving matrix is invertible using the Banach Lemma I have an assignment question that goes like this:
Consider the $n \times n$ matrix
$$
        \begin{pmatrix}
      2 & 1 & 2^{-1} & 2^{-2} & 2^{-3} & 2^{-4} & \cdots & \cdots & 2^{-(n-2)} \\
      2^{-(n-2)} & 2 & 1 & 2^{-1} & 2^{-2} & 2^{-3} & \cdots & \cdots & 2^{-(n-3)} \\
      2^{-(n-3)} & 2^{-(n-2)} & 2 & 1 & 2^{-1} & 2^{-2} & \cdots & \cdots & 2^{-(n-4)} \\
      \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & & & \vdots \\
      \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & & & \vdots \\
      1 & 2^{-1} & 2^{-2} & 2^{-3} & 2^{-4} & 2^{-5} & \cdots & 2^{-(n-2)} & 2
        \end{pmatrix}
$$
Write down this matrix for the case $n = 5$. Use the Banach Lemma (with the matrix infinity-norm) to show that $A_n$ is invertible for any positive integer $n$. Also use the Banach Lemma to derive a suitable upper bound on the infinity-norm of $A_n^{-1}$.
I figured out the matrix for $n= 5$. But after that, I have no idea how to solve this problem. Can someone explain it to me?
Thank you.

Scale the matrix by one half. The new matrix is less than one away from the identity matrix. Apply Banach and undo the scaling.
EDIT: Let $A$ denote your matrix. Then $S = \frac{1}{2}A$ has ones on the main diagonal. Moreover, along each row the off diagonal entries are the just the numbers
\begin{equation}
2^{-1}, 2^{-2}, \dotsc,  2^{-(n-1)}
\end{equation}
in some order. These numbers sum to $1 - 2^{-(n-1)}$ so 
\begin{equation}
\|S-I\|_\infty = 1 - 2^{-(n-1)} < 1.
\end{equation}
Now this is the critical point: By writing $S = I + (S-I)$ you can regard $S$ as a modest perturbation of the identity matrix. By Banach's lemma $S$ is non-singular and you can write
\begin{equation}
\|S^{-1}\|_\infty \leq \frac{1}{1 - \|I-S\|_\infty} = \frac{1}{1 - (1 - 2^{-(n-1)})} = 2^{n-1}.
\end{equation}
Since $A = 2S$ we have that $A$ is nonsingular, and an estimate of $\|A^{-1}\|_\infty$ follows quickly.
