Is there anyway to get eigenvalues or some sort of information about eigenvalues of below matrix? For example, $x^TAx<0$ (hence all eigenvalues are negative) whenever $b_i\leq a_i ~\forall i$.
\begin{bmatrix} 0 & b_2 & b_3 & \cdots & b_{n-1} & b_n \\ -a_2 & -c_2 & 0 & \cdots & 0 & 0 \\ -a_3 & 0 & -c_3 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_n & 0 & 0 & \cdots & 0 & -c_{n} \\ \end{bmatrix} Where $a_i, b_i$ and $c_i$, all are non-negative. Your help will be appreciated.
I don't recall any closed-form formula for the eigenvalues of $A$, but you may use Gerschgorin disc theorem to help locating the eigenvalues. There are stronger versions of the theorem that give more precise locations and you may go to look them up in reference books.
By the way, the sign of $x^TAx$ is equal to the sign of $x^TSx$, where $S$ is the symmetric part of $A$. So, it doesn't matter whether $b_i-a_i$ is negative or positive, because the matrices $\pmatrix{0&\frac12(b-a)\\ \frac12(b-a)&-c}$ and $\pmatrix{0&\frac12(a-b)\\ \frac12(a-b)&-c}$ are always similar to each other. Also, if some $a_i\ne b_i$, e.g. when $a_2\ne b_2$, it is impossible that $x^TAx$ is always nonnegative, because the $2\times2$ leading principal minor of the symmetric part of $A$, i.e. $\pmatrix{0&\frac12(b_2-a_2)\\ \frac12(b_2-a_2)&-c}$, must have a positive eigenvalue.