Is there an easy way of determining if the eigenvalues of a realvalued reverse bidiagonal matrix are real. Basically I have two vectors $(a_1,...,a_n)$ and $(b_1,...,b_{n1})$ that form the "reverse" diagonals of a matrix A. So that $A_{1,n}=a_1, ..., A_{n,1}=a_n$ and $A_{1,n1}=b_1,...,A_{n1,1}=b_{n1}$ 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 $b_i b_{ni} \ge 0$ for $1 \le i < n/2$ and $a_i a_{n+1i} \ge 0$ 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 $a_i a_{n+1i} < 0$ and $b_i, b_{i1}, b_{n+1i}, b_{ni}$ are sufficiently small, there are nonreal eigenvalues (close to $\pm \sqrt{a_i a_{n+1i}}$). Similarly if $b_i b_{ni} < 0$ and $a_i, a_{i+1}, a_{ni}, a_{n+1i}$ are sufficiently small, there are nonreal eigenvalues close to $\pm \sqrt{b_i b_{ni}}$. 

