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Let be $A$ a matrix, $n\times n$ with integer coordinates, and such that $A$ have an eigenvector $\underline{n}$ with real coordenates and eigenvalue 1. My problem is: how to show that $ A $ has an eigenvector with integer coordinates (with eigenvalue 1)

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If we bring $A-I$ to reduced row echelon form (RREF) then the entries will be rational. From the RREF we can read off an eigenvector with rational coordinates. Since any multiple of an eigenvector is still an eigenvector corresponding to the same eigenvalue, it follows that we can clear denominators to produce an eigenvector with integer coordinates.

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Forgive me if I'm ignorant, but I do not quite understand your answer, would you be more clear? –  user27456 Sep 25 '12 at 22:40
    
Which part is confusing to you? –  EuYu Sep 25 '12 at 22:43

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