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For a real symmetric matrix $M$ and a simple eigenvalue $\lambda$, under which conditions the corresponding eigenvector has no zero entries? Perhaps, this is unconditional and one can provide a proof?

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    $\begingroup$ Certainly no-$0$-entry-ness is not unconditional; consider a diagonal matrix with distinct entries. $\endgroup$
    – LSpice
    Commented Oct 26, 2022 at 18:20

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Any $n$ orthogonal vectors are eigenvectors of some symmetric matrix.

One example of a sufficient condition which implies that all coordinates of an eigenvector are non-zero is that the matrix has positive entries, and the eigenvector corresponds to the largest eigenvalue (Perron-Frobenus theorem). In this case, the eiegnvalue of largest absolute value is positive and simple, and the corresponding eigenvectors can be chosen to have all positive coordinates.

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  • $\begingroup$ Thx to both of you. I am working on this, and at this moment it seems that the $(i, j)$-entry is non-zero iff the submatrix obtained by deleting the $i$th row and the $j$th column has no $\lambda$ as an eigenvalue. $\endgroup$
    – Vladimir
    Commented Oct 26, 2022 at 19:15

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