Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
1  
Typo in the title; typo in your final inequality –  Yemon Choi May 10 '12 at 8:14
    
Can you do this for $2\times 2$ matrices? –  Yemon Choi May 10 '12 at 8:14
    
It seems wrong: take just A=diag(10+epsilon, 1) B=diag(10,2) –  Alexander Chervov May 10 '12 at 8:19
    
Alexander, A-B=diag(epsilon,-1) in your example so it doesn't work. –  Felix Goldberg May 10 '12 at 8:48
    
@Felix, yeh, you right, I misread as abs(Re(Lambda_min))... –  Alexander Chervov May 10 '12 at 9:16

1 Answer 1

up vote 2 down vote accepted

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!

share|improve this answer
    
(@Felix: For the matrix display: you have to use FOUR backslashes instead of 2...) –  Tom De Medts May 10 '12 at 9:07
    
@ Tom De Medts, thanks, this is great! –  Felix Goldberg May 10 '12 at 9:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.