Sign pattern symmetric matrices I am interested in sign pattern symmetric real matrices ($a_{ij} a_{ji} \ge 0$ for all $i \ne j$).
I have seen a published proof that such sign-symmetric matrices cannot have purely imaginary eigenvalues.  However, I have a suspicion there are problems with the proof and actually suspect that the phenomenon is not true.
I would appreciate advice.
 A: A trivial counterexample is a zero matrix of any order, but I shall assume that by "purely imaginary" you imply that it is also non-zero.
By just randomly typing matrices and evaluating their eigenvalues, I got this:
$$X = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 5 \\ 1 & 2 & 4 & 8 \end{bmatrix}.$$
The eivenvalues of $X$ are approximately:
$$\lambda_1 \approx 12.7196, \quad \lambda_2 \approx 0.686888 + 0.608444 i, \quad \lambda_3 \approx 0.686888 - 0.608444 i, \quad \lambda_4 \approx -0.0933694.$$
None of these are, of course, purely imaginary, but this is not a problem. Just observe $X' := X - \operatorname{Re}(\lambda_2) I$. Obviously, $X'$ has the same off-diagonal elements as $X$, so it is sign-symmetric, but it also has eigenvalues $\lambda'_k = \lambda_k - \operatorname{Re}(\lambda_2)$. Specifically, this means that $\lambda'_2$ and $\lambda'_3$ are purely imaginary.
Notice that if your conjecture was true, this would mean (from my construction above) that all sign-symmetric matrices have strictly real eigenvalues, which would be a very nice result that someone would have probably noticed by now and we'd all know it, just like we know that this is true for all symmetric and Hermitian matrices.
You should probably check your definition of sign-symmetry. I have found one in this paper and it is far from simply comparing signs on the symmetric positions. I didn't read paper beyond intro, but a quick search revealed no occurrences of the world "imaginary", even though the definition of sign-symmetry there is stricter than your (it includes your criterion plus a similar one for all pairs of symmetrically positioned square submatrices).
A: Perhaps you remember the wrong statement. I guess what you read is actually the following, which is a well known result:

Every real tridiagonal matrix $A$ that is sign-symmetric has only real eigenvalues.

Edit. When all entries in both the superdiagonal and subdiagonal of $A$ are nonzero, the above statement holds because $DAD^{-1}$ is real symmetric for some diagonal matrix $D$. When some $a_{i,i+1}$ or $a_{i+1,i}$ are zero, the statement is still true because matrix eiegnvalues vary continuously with matrix entries.
