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Is there an easy way of determining if the eigenvalues of a real-valued reverse bidiagonal matrix are real. Basically I have two vectors $(a_1,...,a_n)$ and $(b_1,...,b_{n-1})$ that form the "reverse" diagonals of a matrix A. So that $A_{1,n}=a_1, ..., A_{n,1}=a_n$ and $A_{1,n-1}=b_1,...,A_{n-1,1}=b_{n-1}$ and all other $A_{i,j}=0$. What conditions must the vectors $a$ and $b$ satisfy to guarantee that $A$ has real eigenvalues? Thanks for help.

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One sufficient condition is that $b_i b_{n-i} \ge 0$ for $1 \le i < n/2$ and $a_i a_{n+1-i} \ge 0$ for $1 \le i < (n+1)/2$. In fact, if $\ge$ is replaced by $>$, there is a diagonal matrix $U$ such that $U A U^{-1}$ is symmetric.

On the other hand, if $a_i a_{n+1-i} < 0$ and $|b_i|, |b_{i-1}|, |b_{n+1-i}|, |b_{n-i}|$ are sufficiently small, there are non-real eigenvalues (close to $\pm \sqrt{a_i a_{n+1-i}}$). Similarly if $b_i b_{n-i} < 0$ and $|a_i|, |a_{i+1}|, |a_{n-i}|, |a_{n+1-i}|$ are sufficiently small, there are non-real eigenvalues close to $\pm \sqrt{b_i b_{n-i}}$.

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