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Is the maximum eigenvalue (or spectral radius) of the matrix with the following form equalled to row or column sum of the matrix?

$$ A=\left( \begin{array}{cccc} 0 & a & ... & a \\ a & 0 & ... & a \\ : & : & ...& : \\ a & a & ... & 0\end{array} \right) $$

The matrix is square with dimension $n \times n$ where $n = 2,3,4,...$, hollow (all elements in the principal diagonal = 0), symmetric and all off diagonal elements have the same value.

Is the spectral radius of such matrices = $(n-1)\times a$? Why?

Rodrigo de Azevedo
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Val K
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1 Answers1

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Start with the matrix $A$ of all $a$'s, whose eigenvalues are zero except for eigenvalue $na$ having multiplicity one (because rank$(A) = 1$).

Now subtract $aI$ from $A$ to get your matrix. The eigenvalues of $A-aI$ are those of $A$ shifted down by $a$. We get a eigenvalue $(n-1)a$ of multiplicity one and eigenvalue $-a$ with multiplicity $n-1$.

So the spectral radius (largest absolute value of an eigenvalue) of $A$ is $|na|$, and the spectral radius of $A-aI$ is $\max(|(n-1)a|,|a|)$. The latter is simply $|(n-1)a|$ unless $n=1$.

hardmath
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    For $A$ of all $a$'s, if we get n-1 eigenvalues = zero, and one eigenvalue = na, then shouldn't the eigenvalues of the "shifted" matrix $A-aI$ being $-a$ with multiplicity $(n-1)$ and $(n-1)a$ with multiplicity 1? – Val K Mar 28 '17 at 16:10
  • I was mistaken in my subtraction! See my edit. – hardmath Mar 28 '17 at 16:16