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I am trying to calculate the eigenvalues of a square $n \times n$ matrix whose entries are $0$ along the main diagonal and $1$ elsewhere. How do I compute the eigenvalues of this matrix?

I'm thinking along the lines of diagonalisation, but I am not sure at this rate.

Rodrigo de Azevedo
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kevinl8
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  • See circulant matrix. It will also be helpful if you mention the background question you are trying to solve, it will serve as context and it is possible that there may be approaches to that question which could go via a different route. – Sarvesh Ravichandran Iyer Aug 02 '21 at 04:43
  • This is to find the eigenvalues of the complete graph, Kn. I am trying to compute the spectrum. – kevinl8 Aug 02 '21 at 04:55
  • I see, Kevin! Yes, then the matrix you've written down is the correct one to consider, and the matrix that you have written is a special case of a circulant matrix. Basically, note that the matrix $M$ that you have, satisfies some row/column cyclicity properties. If you write these cyclicity properties in terms of matrices, then you'll see where roots of unity come in, and subsequently you can answer the question using those quantities. That's basically what occurs with circulant matrices. – Sarvesh Ravichandran Iyer Aug 02 '21 at 05:04
  • Related: https://math.stackexchange.com/q/904926/339790 and https://math.stackexchange.com/q/84206/339790 – Rodrigo de Azevedo Aug 02 '21 at 11:13

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Note: $$\begin{vmatrix}-t & 1 & \cdots &1&1\\ 1 & -t & \cdots & 1& 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1& 1 &\cdots & -t& 1\\ 1& 1 &\cdots & 1& -t\end{vmatrix}=0 \stackrel{\text{add columns}}{\Rightarrow} \\ (n-1-t)\begin{vmatrix}1 & 1 & \cdots &1&1\\ 1 & -t & \cdots & 1& 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1& 1 &\cdots & -t& 1\\ 1& 1 &\cdots & 1& -t\end{vmatrix}=0 \stackrel{R_1-R_i\to R_i,i=2,3,...,n}{\Rightarrow} \\ (n-1-t)\begin{vmatrix}1 & 1 & \cdots &1&1\\ 0 & 1+t & \cdots & 0& 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0& 0 &\cdots & 1+t& 0\\ 0& 0 &\cdots & 0& 1+t\end{vmatrix}=0 \Rightarrow \\ (n-1-t)(1+t)^{n-1}=0 \Rightarrow \\ t_1=n-1,t_i=-1, i=2,3,...,n.$$

farruhota
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Consider the $n\times n$ matrix $J$ with all entries equal to $1$, with $n>1$. Then its rank is $1$ and $n$ is an eigenvalue. So the algebraic multiplicity of the eigenvalue $0$ is at most $n-1$, but its geometric multiplicity is $n-1$. Hence we know all the eigenvalues with their algebraic multiplicity.

Hence the characteristic polynomial $p_J(x)$ is $(n-x)(0-x)^{n-1}$, by definition of algebraic multiplicity. Therefore $$ \det(J-xI)=(n-x)(0-x)^{n-1} $$ and your matrix is obtained with $x=1$, so the determinant is $(-1)^{n-1}(n-1)$

egreg
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