# On an eigenvalue inequality [closed]

Let $\lambda_1 (\cdot)$ be the larger absolute value eigenvalue of a $2\times2$ matrix and $\lambda_2 (\cdot)$ the smaller absolute value eigenvalue of a $2\times2$ matrix, i.e. $|\lambda_1 (\cdot)| \ge |\lambda_2 (\cdot)|$. Is it true that $$\Big||\lambda_1 (A+B)|-|\lambda_1 (A)|\Big|^{1/3}+\Big||\lambda_2 (A+B)|-|\lambda_2 (A)|\Big|^{1/3}\leq|\lambda_1 (B)|^{1/3}+|\lambda_2 (B)|^{1/3}$$ for any two $2\times2$ symmetric real matrices $A$ and $B$? Thanks a lot!

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## closed as no longer relevant by Andrés Caicedo, Andy Putman, Suvrit, Will Jagy, Harald Hanche-OlsenFeb 21 '12 at 15:33

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You might give a link to the m.se question so we could easily check what's there and not duplicate work. – Gerry Myerson Dec 28 '11 at 2:10
Where does this question come from? – Igor Rivin Dec 28 '11 at 11:59
Firstly: since Suvrit answered your original question, why not accept his answer? – Yemon Choi Dec 28 '11 at 19:47
@unknown (yahoo): my sincere apologies for misreading your new question in haste. [earlier comment now deleted] (But the restriction that $B$ be diagonal real, rather than merely real-symmetric, still seems superfluous.) – Yemon Choi Dec 28 '11 at 20:30
In fact, I suggest you post the new question as a separate question, but include a link back to this one in order to provide background context – Yemon Choi Dec 28 '11 at 20:32

The alleged inequality is false, even if you restrict $A$ and $B$ to be positive definite matrices. Consider the following,
$$A = \begin{bmatrix} 1.2281 & 0.6361\\\\ 0.6361 & 1.9690 \end{bmatrix},\quad\quad B = \begin{bmatrix} 3.7829 &-0.6021\\\\ -0.6021 & 0.4002 \end{bmatrix}.$$ Then, we have the following: