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.

  • $\begingroup$ @MorganRodgers but getting $n$ degree polynomial, can we get any kind of information about eigenvalues? your introduction of $-c_1$ is good idea. appreciated!! $\endgroup$ Aug 2, 2016 at 16:13
  • $\begingroup$ @RealHilbert each coefficient of the characteristic polynomial tells you something about the eigenvalues (e.g. the determinant and trace). Also, when you say "the eigenvalues are negative", do you mean that the eigenvalues have a negative real part? $\endgroup$ Aug 2, 2016 at 19:34
  • $\begingroup$ @Omnomnomnom eigenvalues are negative only not negative real part $\endgroup$ Aug 3, 2016 at 15:48
  • $\begingroup$ Then your supposition that $A$ will have negative eigenvalues is wrong. $\endgroup$ Aug 3, 2016 at 17:29
  • $\begingroup$ @Omnomnomnom How? $\endgroup$ Aug 3, 2016 at 17:44

1 Answer 1


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.

  • $\begingroup$ so you are saying that if consider the case $n=2$, then matrix will have a positive eigenvalue. That is not true. $\endgroup$ Aug 3, 2016 at 15:46
  • $\begingroup$ the matrix I have posted in the question. $\endgroup$ Aug 3, 2016 at 16:49
  • $\begingroup$ is it a theorem? but $x^TAx<0 ~~\forall x$ whenever $b_2 \leq a_2 $. Can you give some references ? $\endgroup$ Aug 3, 2016 at 17:13
  • 1
    $\begingroup$ @RealHilbert Your assertion isn't true. A simple counterexample is $(-1,1)\pmatrix{0&1\\ -4&0}\pmatrix{-1\\ 1}=3>0$. In general, $x^TAx=x^T\frac{A+A^T}2x$. So, it suffices to consider only the symmetric part $S$ of $A$. Now, if $a_2\ne b_2$, the leading principal $2\times2$ submatrix of $S$ -- say $S_2$ -- would have a negative determinant. Since the two eigenvalues of $S_2$ are real, one of them must be positive. Let $v\in\mathbb R^2$ be an eigenvector for this positive eigenvalue. Then $(v^T,0)A\pmatrix{v\\ 0}=(v^T,0)S\pmatrix{v\\ 0}=v^TS_2v>0$. $\endgroup$
    – user1551
    Aug 3, 2016 at 17:23
  • $\begingroup$ my bad I assumed $c_1>0$. but still what you claiming that should be true in the case $c_1>0$ also?? $\endgroup$ Aug 3, 2016 at 17:36

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