# Eigenvalue of diagonally dominant matrices

I want to ask a question about eigenvalue of diagonally dominant matrices.

The question is :

Assume $$A=(a_{ij})_{n\times n}\in M_n(\mathbb{R})$$ and $$\lambda_1,\lambda_2, \cdots,\lambda_n$$ are the eigenvalues of $$A$$. $$A$$ satisfies:

$$(1):$$ $$a_{ii}>0~~,i=1,2,\cdots,n$$.

$$(2):$$ $$a_{ij}<0~~,i\ne j$$.

$$(3):$$ $$a_{ii}>-\sum_{i\ne j}a_{ij}~~,i=1,2,\cdots,n$$.

I would like to ask if there is $$\mathrm{Re}(\lambda_i)\ge\vert \mathrm{Im}(\lambda_i)\vert,~~ i=1,2,\cdots,n$$ ?

I think it might have something to do with the estimation of the eigenvalues of the diagonal dominance matrix. Any help or suggestions are appreciated.

Maybe it is not true and I can't find an counterexample.

Thanks!

• It would be true if you had $a_{ii}>-2\sum_{i\ne j}a_{ij}$ using the Gerschgörin circle theorem, in case that's useful to you, but with only your condition there might be counterexamples. That's only my guess though. Commented Jul 26, 2023 at 6:16

Choose an $$n \geq 3$$, and consider the matrix $$A=\begin{bmatrix} 1& -1& 0& \cdots & 0 \\ 0 & 1 & -1 & 0 & \cdots \\ \cdots & & & &\\ -1 & 0&0&\cdots&1 \end{bmatrix}$$ The matrix has characteristic equation $$(1 -x)^n=1$$ which implies that it’s eigenvalues are $$1-z_n^i$$ for the roots of unity $$z_n$$.

Now consider the matrix $$A-\epsilon J+ 3n\epsilon I$$. Here J is the all-one matrix. For sufficiently small $$\epsilon>0$$, this matrix satisfies all constraints to the problem. Furthermore, its spectrum converges to the spectrum of $$A$$ as $$\epsilon$$ goes to zero.

In particular, even for $$n=6$$ setting $$\epsilon$$ sufficiently close to zero gives a counter example for your conjecture given by $$i=2$$.

Exercise: Prove that the set of possible eigenvalues of matrices satisfying your constraint is precisely the right half plane.

• In your example, $a_{ii}\not>-\sum_{i\ne j}a_{ij}~~,i=1,2,\cdots,n$. Commented Aug 2, 2023 at 14:35
• Sure, but one can perturb the entries slightly to make this happen, unless I am misunderstanding the problem. Commented Aug 2, 2023 at 14:39
• Some intuitions may look correct, but are proved false rigorously. Commented Aug 2, 2023 at 14:43
• I made everything completely rigorous. Happy? Commented Aug 2, 2023 at 14:47
It isn’t clear whether the statement you want to prove/disprove is $$\operatorname{Re}(\lambda)\ge|\operatorname{Im}(\lambda)| \ \text{ for all eigenvalues \lambda of A}\tag{1}$$ or $$\operatorname{Re}(\lambda)\ge|\operatorname{Im}(\lambda)| \ \text{ for some eigenvalue \lambda of A.}\tag{2}$$ The answer by abacaba has given a beautiful counterexample to $$(1)$$. Below we shall prove statement $$(2)$$. Let $$D$$ and $$F$$ be the diagonal and off-diagonal parts of $$A$$ respectively. Then $$DF,\,FD$$ are hollow matrices and $$D^2,F^2$$ are entrywise nonnegative. Therefore $$\operatorname{tr}(A^2) =\operatorname{tr}\big((D+F)^2\big) =\underbrace{\operatorname{tr}(D^2)}_{\ge0}+\underbrace{\operatorname{tr}(DF)}_{=0} +\underbrace{\operatorname{tr}(FD)}_{=0}+\underbrace{\operatorname{tr}(F^2)}_{\ge0} \ge0.$$ Hence $$\operatorname{Re}(\lambda^2)\ge0$$ for some eigenvalue $$\lambda$$ of $$A$$, i.e., $$|\operatorname{Arg}(\lambda^2)|\le\frac{\pi}{2}$$. As $$A$$ is a strictly diagonally dominant matrix with a real positive diagonal, we also have $$\operatorname{Re}(\lambda)>0$$. Therefore $$|\operatorname{Arg}(\lambda)|\le\frac{\pi}{4}$$ and $$\operatorname{Re}(\lambda)\ge|\operatorname{Im}(\lambda)|$$.
• I want to prove statement $(1)$. Abacaba has given a beautiful answer. But I still appreciate your proof of statement $2$. Commented Aug 3, 2023 at 3:53