Eigen values of special matrix

Question

I have a matrix $A \in \mathbb{R}^{n\times n}$ whose diagonal elements are all "$1$" and all other elements are of the form $\frac{-1}{n}$ where $n \in \mathbb{N}$ and $n >1$ is the number of rows or columns of $A$. Since $A$ is symmetric and diagonally dominant with positive diagonal entries, it is positive definite. Can we get an expression for its eigenvalues using some kind of decomposition or by other standard approach ?

This matrix clearly has a special structure but I am unable to utilize its properties.

This question (which is addressed in my first link) is a little more specifically geared towards your question. — Ben Grossmann, Aug 07 '17 at 13:27
Long story short: the eigenvalues will be $1/n$ with multiplicity $1$ and $(n+1)/n$ with multiplicity $(n-1)$. — Ben Grossmann, Aug 07 '17 at 13:30

score 1 · Answer 1 · answered Aug 07 '17 at 13:50

Yes! They are always $\dfrac{n+1}{n}$ with multeplicity $(n-1)$ and the "last" is $\dfrac{1}{n}$

$$\frac{n+1}{n},\;\frac{n+1}{n},\ldots,\frac{1}{n}$$

The characteristic polynomial is $$P(\lambda)=\frac{(-1)^n \left((\lambda n-n-1)^{n-1} (\lambda n-1)\right)}{n^n}$$

Determinant is $D=\dfrac{(n + 1)^{n - 1}}{n^n}$

Eigenvectors are special, too

For instance if $n=5$ they are the rows of this matrix $$ \left( \begin{array}{ccccc} -1 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 1 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ \end{array} \right) $$

For $n=6$ the following $$ \left( \begin{array}{cccccc} -1 & 0 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ \end{array} \right) $$

and so on

Hope it helps

Eigen values of special matrix

1 Answers1