What is the determinant value of $J-I$ if $I$ is identity matrix and $J=(1)_{101\times 101}$?

Question

Let $J$ be a matrix of order $101\times 101$ which each entry is 1 and suppose $I_{101}$ is identity matrix of order $101\times 101$. The question is : what should be the determinant value of $J-I$ ?

This one I trying last one week but in vain. What I find is $J-I$ is a matrix which leading diagonal entries are all zero and all other entries as 1. Some thing like this $$J-I=\left(\begin{array}{ccc} 0 & 1 & 1 &\cdots & 1 \\ 1 & 0 & 1 &\cdots & 1 \\ 1 & 1 & 0 &\cdots & 1 \\ & \cdots & & \cdots & \\ 1 & 1 & 1 & \cdots & 0 & \end{array}\right)$$

Its a Toeplitz matrix if I am not wrong. In websites, I have searched how to get the determinant values of such matrix but what I have got is very very complicated idea and I do believe that the solution strategy for this problem would not be that much difficult. But then how to figure it out ?

How to compute its determinant ?

Answer 1

Apply the determinant formula for circulant matrices, found here.

Answer 2

$J$ has eigenvalues $0$ and $101$, with multiplicity $100$ and $1$ respectively. Hence its characteristic polynomial is $p_J(x)=x^{100}(x-101)$. The characteristic polynomial of $J-I$ is $$P_{J-I}(x)=P_J(x+1)=(x+1)^{100}(x-100)$$ Then $$Det(J-I)=(-1)^{101}P_{J-I}(0)=100$$