Suppose $A$ is an $n×n$ symmetric matrix with all entries $0$ or $1$, and with diagonal $0$.
Are all of the eigenvalues of $A$ integers? It works for all the cases I have tried so far.
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asked Feb 8, 2015 at 10:11
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$\begingroup$ en.wikipedia.org/wiki/Integer_matrix "The characteristic polynomial of an integer matrix has integer coefficients. Since the eigenvalues of a matrix are the roots of the polynomial, the eigenvalues of an integer matrix are algebraic integers." $\endgroup$– DemostheneFeb 8, 2015 at 10:22
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2$\begingroup$ @Demosthene yes they will be algebraic numbers by definition, but these can be fractions or even irrational ($\sqrt 2$ is algebraic since it is a root of $x^2-2$). I want the eigenvalues to be actual integers like $1,5,-7$. I know this is not true for integer matrices in general. $\endgroup$– Forever MozartFeb 8, 2015 at 10:26
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2 Answers
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Try $$A=\begin{bmatrix} 0&0&1\\0&0&1\\1&1&0\end{bmatrix}.$$ If I am not wrong its characteristic polynomial is $X(2-X^2)$ and its eigenvalues are $0$, $\sqrt{2}$ and $-\sqrt{2}$.
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Have a look at this counterexample:
