Prove $(I-J_n)^{-1}=I-\frac{1}{n-1}J_n$

Question

I have a question on matrices from a recent problem.

Let $J_n$ be the $n\times n$ matrix each of whose entries is $1$. Show that if $n\gt 1$, then $$(I-J_n)^{-1}=I-\frac{1}{n-1}J_n$$

Here's what I've tried so far:

$J_n$ is a matrix of $1$'s, so $I-J_n$ is the matrix whose diagonal entries are $0$ and $1$ elsewhere. But from here, I don't know how to take the inverse of this $n\times n$ matrix. One could use Gaussian-Jordan elimination but I don't think it helps here, it would get too complicated very quickly. Instead, I called the left-hand side of the equation $K$.

\begin{align} (I-J_n)^{-1} & = K \\ (I-J_n)^{-1}K^{-1} & = KK^{-1} \\ (K(I-J_n))^{-1} & = I \\ (K-KJ_n)^{-1}&=I \\ I & = K-KJ_n \end{align}

I still need to use the fact that $J_n$ contains only $1$'s but I'm not sure how to do that.

Any help in the right direction is appreciated.

Answer 1

Simply test the multiplication $(I - J_n)\left(I - \frac{1}{n - 1}J_n\right)$:

\begin{align} & (I - J_n)\left(I - \frac{1}{n - 1}J_n\right) \\ = & I - \frac{1}{n - 1}J_n - J_n + \frac{1}{n - 1}J_n^2 \\ = & I - \frac{n}{n - 1}J_n + \frac{n}{n - 1}J_n \\ = & I, \end{align} where we used $J_n^2 = nJ_n$, which can be easily verified.

Answer 2

$J_n^2=nJ_n$, so $(I-J_n)\left(I-\dfrac1{n-1}J_n\right)=I-\dfrac1{n-1}J_n-J_n+\dfrac1{n-1}J_n^2=I$