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Let $A$ be an irreducible non-negative matrix. Is it true that the eigenvectors of $A$ can span the $R^n$ ?

Or are all the eigenvalues of $A$ distinct?

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  • $\begingroup$ I guess you seek the property for primitive nonnegative matrices (stronger than irreducible). The Perron Frobenius eigenvalue is then a simple extremal eigenvalue if i am not mistaken. $\endgroup$
    – Toni Mhax
    Commented Dec 23, 2022 at 16:55

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You don't need to search for complicated counterexamples; just consider the matrix with all elements equal to 1.

[EDIT: removed a second counterexample after a comment pointed out it was reducible. If you want an example with all distinct eigenvalues, you can take the cyclic shift matrix.]

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  • $\begingroup$ First example is correct. Second one is reducible i think $\endgroup$
    – Razor
    Commented Dec 23, 2022 at 18:29
  • $\begingroup$ @Razor oops you are correct. Edited. $\endgroup$
    – Federico Poloni
    Commented Dec 23, 2022 at 18:51

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