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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!

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    $\begingroup$ Typo in the title; typo in your final inequality $\endgroup$
    – Yemon Choi
    May 10, 2012 at 8:14
  • $\begingroup$ Can you do this for $2\times 2$ matrices? $\endgroup$
    – Yemon Choi
    May 10, 2012 at 8:14
  • $\begingroup$ It seems wrong: take just A=diag(10+epsilon, 1) B=diag(10,2) $\endgroup$
    – Alexander Chervov
    May 10, 2012 at 8:19
  • $\begingroup$ Alexander, A-B=diag(epsilon,-1) in your example so it doesn't work. $\endgroup$
    – Felix Goldberg
    May 10, 2012 at 8:48
  • $\begingroup$ @Felix, yeh, you right, I misread as abs(Re(Lambda_min))... $\endgroup$
    – Alexander Chervov
    May 10, 2012 at 9:16

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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!

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

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