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Let $i_0\in \{1,\ldots ,n\},\alpha >0$. Then$$K_{i_0}:=\left \{z\in \mathbb{C}:|z-a_{i_0i_0}|\leq \alpha \sum \limits _{j=1 \\ j\neq i_0}^n|a_{i_0j}|\right \}$$has exactly one eigenvalue if and only if$$K_{i_0}\cap \left \{z\in\mathbb{C}:|z-a_{ii}|\leq \frac{|a_{ii_0}|}{\alpha}+\sum \limits _{j=1 \\ j\neq i,i_0}^n |a_{i_0j}|\right \}=\varnothing \quad \forall i\in \{1,\ldots ,n\}\setminus \{i_0\}.$$

I have no idea how to prove this. I know about Gershgorin discs and that all eigenvalues are in the discs and I have found a few proves about the amount of eigenvalues of disjointed discs but not with exactly one eigenvalue and $\alpha$. Thanks for any help!

Uhmm
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  • Uh...the title is really misleading here, enough that I assumed you meant something entirely different until I actually read what you wrote carefully. This is "Gerschgorin-like" but not especially so. The top disk is only a Gerschgorin disk when $\alpha=1$ and even then the bottom disks are not Gerschgorin disks. In any case I am curious what happens with $\begin{bmatrix} 0 & 1/3 \ 2/3 & 1 \end{bmatrix}$, which is a counterexample to the "naive" statement of this type. – Ian May 18 '22 at 19:15
  • Never mind, checking that means this whole thing is just nonsense. Clearly it is possible for $K_{i_0}$ to only contain one eigenvalue, just take a diagonal matrix with distinct eigenvalues. But the second statement is trivial in the 2x2 case because the sum is just empty. So something is going wrong here. – Ian May 18 '22 at 19:22
  • @Ian hmm okay, thank you for your help!! – Uhmm May 18 '22 at 20:17
  • @Ian yeah we just had Gershgorin circles and this seemed to be very close to it, I'm sorry that I mislead you by this title! – Uhmm May 18 '22 at 20:19
  • Assuming this problem was given to you, I wonder what the correct version of the problem actually is. As I said, this version is definitely not correct. – Ian May 18 '22 at 20:32
  • @Ian there was a mistake: I made an edit – Uhmm May 18 '22 at 20:35
  • That's not completely nonsensical like the other version (and is also much more similar to the usual Gerschgorin than the other version). But it still can't be true as is, at least not if $\alpha$ is just whatever positive number you want. Take $\alpha=1$, look at $\begin{bmatrix} 0 & 1/3 \ 2/3 & 1 \end{bmatrix}$ and then consider $i_0=1,i=2$. The disk at the top is $B_{1/3}(0)$, which only contains one eigenvalue, but the disk at the bottom is $B_{2/3}(1)$, and they have a point in common, namely $1/3$. – Ian May 18 '22 at 20:46

1 Answers1

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This just isn't true, at least if $\alpha$ can be anything.

As a whole family of counterexamples, look at $A=\begin{bmatrix} 0 & x \\ 2x & 1 \end{bmatrix}$. The eigenvalues are $\frac{1 \pm \sqrt{1+8x^2}}{2}$ and you have $K_1=B_{\alpha x}(0)$. Say the lower type of disk is denoted $J_{i,i_0}$, then you have $J_{2,1}=B_{2x/\alpha}(1)$. $K_1 \cap J_{2,1}$ is nonempty if $\alpha x \geq 1-2x/\alpha$ i.e. $-1 \geq x(-2/\alpha-\alpha)$ i.e. $x \geq \frac{1}{2/\alpha+\alpha}$.

So now the question is, can that happen while $K_1$ contains exactly one eigenvalue? This amounts to asking for the following three inequalities: $-\alpha x \leq \frac{1-\sqrt{1+8x^2}}{2},x \geq \frac{1}{2/\alpha+\alpha},\frac{1+\sqrt{1+8x^2}}{2}>\alpha x$. For $\alpha=1$ these are all true if $x \in [1/3,1]$.

Ian
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