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Suppose that $A$ is an invertible $3 \times 3$ matrix with integral entries and that $v$ is an eigenvector corresponding to an irrational eigenvalue. Of course by scaling $v$ we may arrange that it has at least one rational entry. But is it possible for $v$ to have two rational entries, and only one irrational one?

(Feel free to share any higher-dimensional results, of course!)

Tim Piyim
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1 Answers1

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$$ \pmatrix{0&2\\ 1&0\\ &&I_{n-2}}\pmatrix{\sqrt{2}\\ 1\\ \mathbf0}=\sqrt{2}\pmatrix{\sqrt{2}\\ 1\\ \mathbf0}. $$

user1551
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