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Define $Re\lambda_{min}(A)$ to be the minimum of the real parts of the eigenvalues of a matrix $A$. Let $A,B\in \mathbb{R}^{n \times n}$ be two matrice such that $Re\lambda_{min}(A)>0$, $Re\lambda_{min}(B)>0$ and $Re\lambda_{min}(A-B)\geq 0$. Is $Re\lambda_{min}(A)\geq Re\lambda_{min}(B)$ right? And how to prove.

Any help will be appreciated!

edited May 10 2012 at 10:14

asked May 10 2012 at 8:05

W. Nyway
55●3

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Typo in the title; typo in your final inequality – Yemon Choi May 10 2012 at 8:14

Can you do this for $2\times 2$ matrices? – Yemon Choi May 10 2012 at 8:14

It seems wrong: take just A=diag(10+epsilon, 1) B=diag(10,2) – Alexander Chervov May 10 2012 at 8:19

Alexander, A-B=diag(epsilon,-1) in your example so it doesn't work. – Felix Goldberg May 10 2012 at 8:48

@Felix, yeh, you right, I misread as abs(Re(Lambda_min))... – Alexander Chervov May 10 2012 at 9:16

Felix Goldberg · Accepted Answer · 2012-05-10 09:11:14Z UTC

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Here is a $2 \times 2$ counterexample:

$A=\begin{bmatrix}10 & 19 \\ 8 & 16\end{bmatrix}$

$B=\begin{bmatrix}9 & 2 \\ 17 & 7\end{bmatrix}$

If you have tacit assumptions, share them and we'll see if the statement becomes true!

edited May 10 2012 at 9:11

answered May 10 2012 at 9:02

Felix Goldberg
2,383●4●12

	(@Felix: For the matrix display: you have to use FOUR backslashes instead of 2...) – Tom De Medts May 10 2012 at 9:07
	@ Tom De Medts, thanks, this is great! – Felix Goldberg May 10 2012 at 9:11

Inequality regarding the smallest real part of eigenvales

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