Wednesday, December 26, 2018

linear algebra - Maximum eigenvalue of a hollow symmetric matrix


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?


Answer



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 $n=1$.


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