Inverse of matrix of ones minus identity matrix

Question

What is the solution for

$$A^{-1}(n) = (\mathbf{1}_{n} - I_{n})^{-1},$$ where $\mathbf{1}_{n}$ is the $n\times n$ matrix of ones and $I_{n}$ is the $n\times n$ identity matrix.

Numerical Examples Suggest

$$ A^{-1}(n) = \begin{bmatrix} \frac{-n+2}{n-1} & \frac{1}{n-1} & \frac{1}{n-1} & ... \\ \frac{1}{n-1} & \frac{-n+2}{n-1} & \frac{1}{n-1} & ... \\ \frac{1}{n-1} & \frac{1}{n-1} & \frac{-n+2}{n-1} & ... \\ \vdots & \vdots & \vdots & \frac{-n+2}{n-1} \\ \end{bmatrix} $$

For similar questions on determinants, see

How to calculate the determinant of all-ones matrix minus the identity?

Why is the determinant of the all one matrix minus the identity matrix n-1?

How to calculate the determinant of all-ones matrix minus the identity?

Determinant of a matrix with diagonal entries $a$ and off-diagonal entries $b$

Determinant of a specific circulant matrix, $A_n$

Answer 1

This is a direct application of the Sherman–Morrison formula

$$ (A+uv^{T})^{-1}=A^{-1}-{A^{-1}uv^{T}A^{-1} \over 1+v^{T}A^{-1}u} $$

with $A=-I_n$ (where $I_n$ is the identity matrix), $v=w=\mathbf{1_1}$ (vectors filled with $1$)

Answer 2

Here is a very elementary approach : by direct computation, $A$ satisfies $$A^2 =(n-2)A + (n-1)I_n$$

One deduces that $$A \times \left[\dfrac{1}{n-1}\Big(A + (2-n)I_n \Big) \right] = I_n,$$ and therefore, $A$ is invertible and $$\boxed{A^{-1} = \dfrac{1}{n-1}\left(A + (2-n)I_n \right)}$$