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 $$ \left|\left|\lambda_{1}\left(A+B\right)\right|^{1/3}-\left|\lambda_{1}\left(A\right)\right|^{1/3}\right|+\left|\left|\lambda_{2}\left(A+B\right)\right|^{1/3}-\left|\lambda_{2}\left(A\right)\right|^{1/3}\right|\leq\left|\lambda_{1}\left(B\right)\right|^{1/3}+\left|\lambda_{2}\left(B\right)\right|^{1/3} $$ for any $2\times2$ symmetric real matrix $A$ (would suffice to prove or disprove for not positive-definite matrices $A$) and $2\times2$ diagonal real matrix $B$? Thanks a lot for any helpful answers! By the way, a relevant question was answered by Suvrit here.

  • $\begingroup$ Where do you get these statements from? Suvrit gave a counterexample to a previous claim, so unless there is a good reason to believe this is true, it is not clear that it is worthwhile thinking about these questions. $\endgroup$
    – Igor Rivin
    Dec 28, 2011 at 20:52
  • 1
    $\begingroup$ I think that this inequality does hold (even for $n \times n$ matrices), but I know the proof only for positive definite matrices, which implies a restricted version of the inequality that you actually seem to be after. $\endgroup$
    – Suvrit
    Dec 28, 2011 at 20:55
  • 4
    $\begingroup$ META: tea.mathoverflow.net/discussion/1187/… $\endgroup$
    – Will Jagy
    Dec 29, 2011 at 6:19

1 Answer 1


Below I highlight that a much more general claim holds for $n\times n$ positive definite matrices, and that a slightly weaker version of your inequality holds for general symmetric matrices.

Recall a classic theorem of Ando, (T.Ando, "Comparison of norms $\|f(A)-f(B)\|$ and $\|f(|A-B|)\|$, Math. Z., 197, (1988)):

Theorem (Ando). Let $A$ and $B$ be positive semidefinite matrices, and let $\|\cdot\|$ be any unitarily invariant norm, and let $|X| = (X^TX)^{1/2}$ denote the matrix absolute value. For any nonnegative operator monotone function $f(t)$ on $[0,\infty)$,\begin{equation*} \|f(A)-f(B)\| \le \|f(|A-B|)\|\end{equation*}

Now, in your case we can use $f(t) = t^r$ for $r \in [0,1]$, to obtain $$\|A^r-B^r\| \le \|\ |A-B|^r\ \|,$$ which when specialized to the trace-norm (sum of singular values) yields the inequality that you desire (but for positive matrices).

This inequality immediately implies the following weaker one for general symmetric matrices $$\| f(|A|) - f(|B|) \| \le \left\|f\bigl(\bigl|\ |A|-|B|\ \bigr|\bigr)\right\|,$$ which is somewhat weaker than what you desire (but may suffice for your needs---which can be elaborated upon only if you follow Igor's suggestion and tell us where you are getting these questions from, and in what context!)

  • $\begingroup$ In that case, I'm curious to see your proof of the positive definite case! $\endgroup$
    – Suvrit
    Dec 29, 2011 at 12:01
  • $\begingroup$ @unknown: it actually does follow because of the following theorem: $\|\sigma^\arrowdown(f(A)) - \sigma^\arrowdown(f(B))\| \le \|f(A)-f(B)\|$ $\endgroup$
    – Suvrit
    Jan 1, 2012 at 11:08

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