## 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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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 2011 at 2:10
Where does this question come from? – Igor Rivin Dec 28 2011 at 11:59
Firstly: since Suvrit answered your original question, why not accept his answer? – Yemon Choi Dec 28 2011 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 2011 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 2011 at 20:32

## closed as no longer relevant by Andres Caicedo, Andy Putman, S. Sra, Will Jagy, Harald Hanche-OlsenFeb 21 2012 at 15:33

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:

\begin{eqnarray*} \lambda(A+B) = (5.0114, 2.3687)\\ \lambda(A) = (2.3347, 0.8624)\\ \lambda(B) = (3.8868, 0.2962) \end{eqnarray*}

From, which we see that

\begin{eqnarray*}|\ \ |\lambda_1(A+B)| -|\lambda_1(A)|\ \ |^{1/3} + |\ \ |\lambda_2(A+B)|-|\lambda_2(A)|\ \ |^{1/3} & = & 2.5348\\ |\lambda_1(B)|^{1/3} + |\lambda_2(B)|^{1/3} &=& 2.2389 \end{eqnarray*}

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you're unknown, but you're welcome :-) !! – S. Sra Dec 28 2011 at 21:18