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Suppose $M$ is a $n \times n$ matrix such that, $$M=\left[\begin{array}{ccccccc} 0 & k_2&k_3&\cdots &k_{n-1}&k_n\\ k_{1} & 0 & k_3&\cdots &k_{n-1}&k_n\\ k_1&k_{2} & 0 &\cdots &k_{n-1}&k_n\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ k_{1} & k_{2}& k_3&\cdots & 0 &k_n\\ k_1&k_{2} & k_3&\cdots &k_{n-1}& 0\\ \end{array}\right]$$ with the characteristic polynomial, $0=|M-xI|=(-1)^n[x^n - (\displaystyle\sum_{\stackrel{i_1<i_2}{i_1,i_2=1}}^{n}{k_{i_1} k_{i_2}}) \ x^{(n-2)}- (2\displaystyle\sum_{\stackrel{i_1<i_2<i_3}{i_1,i_2,i_3=1}}^{n}{k_{i_1} k_{i_2}k_{i_3}}) \ x^{(n-3)} \ - \ ,\cdots, -(n-2)\displaystyle\sum_{\stackrel{i_1<i_2<i_3<...<i_{n-1}}{i_1,i_2,i_3,...,i_{n-1}=1}}^{n}{k_{i_1} k_{i_2}k_{i_3},...,k_{i_n}}\ x^{(1)}-(n-1)\displaystyle\sum_{\stackrel{i_1<i_2<i_3<...<i_n}{i_1,i_2,i_3,...,i_n=1}}^{n}{k_{i_1} k_{i_2}k_{i_3},...,k_{i_n}}\ x^{(0)} ].$
Then, how to prove that for $k_i\ne k_j$ for some $1\leq i,j\leq n$, the characteristic polynomial $|M-xI|=0$ has at least one roots which is not integer.
(Note that for $k_i = k_j$ for all $1\leq i,j\leq n$, the characteristic polynomial $|M-xI|=0$ has $n$ roots which are integers.)

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It's not true. Any $M$ with exactly one nonzero column can serve as a counterexample. Counterexample exists even if all $k_i$s are nonzero and distinct. E.g. $M=\pmatrix{0&4\\ 1&0}$.

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  • $\begingroup$ you missed the condition that all the entries must be same in $M$ to get integral eigenvalues. $\endgroup$ – L S B. user255259 Sep 23 '15 at 7:41
  • $\begingroup$ @LSB.user255259 What do you mean? Didn't you explicitly specified that the $k_i$s are not all the same? $\endgroup$ – user1551 Sep 23 '15 at 8:01
  • $\begingroup$ If $M$ is $2 \times 2$ matrix with $k_i \times k_j$ is a perfect square then all the eigenvalues are integral. $\endgroup$ – L S B. user255259 Sep 23 '15 at 13:29
  • $\begingroup$ @LSB.user255259 Yes. So this answers your question, doesn't it? You asked how to prove that $M$ has an non-integer eigenvalue when some $k_i$s are distinct, and I showed you a counterexample that all eigenvalues of $M$ can be integers even if all $k_i$s are distinct. $\endgroup$ – user1551 Sep 23 '15 at 13:45
  • $\begingroup$ how to prove it for general $n \times n$. Do you mean that general $n \times n$ matrix for distinct $k_i$ have an integral eigenvalues? as in the case of $2 \times 2$ to get the integral eigenvalues, $k_i$ s must be a perfect square, then what must be the $k_i$s values for $n \times n$ matrix to get the integral eigenvalues. $\endgroup$ – L S B. user255259 Sep 23 '15 at 14:26

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