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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 ?

Martin Sleziak
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KON3
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  • well using Mathematica i get 100 as the answer. – Hubble07 Aug 20 '14 at 07:14
  • Will it be possible to evaluate it manually ? After all, now we know that the answer is 100. – KON3 Aug 20 '14 at 07:15
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    Start with the $2\times2$ version of the problem, then the $3\times3$, then the $4\times4$; guess a pattern, and try to prove it by induction. – Gerry Myerson Aug 20 '14 at 07:16
  • It looks like the general answer for $n \times n$ is $(n-1)(-1)^{n+1}$ – Hubble07 Aug 20 '14 at 07:35
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    See also: http://math.stackexchange.com/questions/84206/how-to-calculate-the-following-determinants-all-ones-minus-i or http://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix (And plenty of other posts, for example, the posts listed there among linked questions.) – Martin Sleziak Aug 20 '14 at 08:34
  • Martin Sleziak, Claude Leibovici, Jyrki Lahtonen, Jack D'Aurizio, Davide Giraud thanks to all of you for sharing the link. That was really helpful. – KON3 Aug 20 '14 at 10:25

2 Answers2

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$J$ 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 $$Det(J-I)=(-1)^{101}P_{J-I}(0)=100$$

Capublanca
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0

Apply the determinant formula for circulant matrices, found here.

Potato
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