6
$\begingroup$

For positive numbers $a$ and $b$ we have the inequality $\sqrt{a+b} \leqslant \sqrt{a} + \sqrt{b}$. Is it true that the same holds if we take $a$ and $b$ to be positive semidefinite matrices?

If not, there is a weaker statement that I am interested in: Is it true that the inequality $\sigma_n( \sqrt{A} - \sqrt{B}) \leqslant \sigma_n(|A-B|^{\frac{1}{2}})$, where by $\sigma_n$ I denote the $n$-th singular value (they are listed in decreasing order), holds?

If even this fails, maybe the following is true: $\sigma_{2n}(\sqrt{A} - \sqrt{B}) \leqslant \sigma_{n} (|A-B|^{\frac{1}{2}})$?

I suspect that analogous inequalities should hold for any $0<r<1$.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

The claim is false. Just try some random psd matrices $A$ and $B$. You can get $\sigma_n( |A-B|^{1/2}) = \sigma_n^{1/2}(A-B) = 0$, whereas $\sigma_n(A^{1/2}-B^{1/2}) > 0$.

Here is an explicit example:

\begin{equation*} A = \begin{pmatrix} 19 & 17 & 9\\ 17 & 17 & 11\\ 9 & 11 & 11\end{pmatrix},\quad B = \begin{pmatrix}19 & 11 & 21\\ 11 & 9 & 15\\ 21 & 15 & 27\end{pmatrix},\quad A-B = \begin{pmatrix}0 & 6 & -12\\ 6 & 8 & -4\\ -12 & -4 & -16\end{pmatrix}. \end{equation*} Now, $\sigma_n(\sqrt{A}-\sqrt{B}) = 0.1853...$, while $\sigma_n(|A-B|^{1/2})=0$.

However, a weaker claim that holds is described in this MO post, namely a weak majorization relation:

\begin{equation*} \|f(A) - f(B)\| \le \|f(|A-B|)\|, \end{equation*} for any symmetric (i.e., unitarily invariant) norm $\|\cdot\|$ and where $f(t) = t^r$, for $0< r < 1$, and more generally, $f$ is a nonnegative concave function.

$\endgroup$
4
  • $\begingroup$ Suvrit, thank you for this counterexample. Unfortunately, this weak majorization would not be useful for the application that I have in mind. $\endgroup$ Jan 17, 2015 at 14:21
  • $\begingroup$ @MateuszWasilewski: Lower bounds on $\sigma_n$ are just too useful to be easily had :-) --- even the majorization result that I cited was quite nontrivial and took a few years of effort to get proved! $\endgroup$
    – Suvrit
    Jan 17, 2015 at 14:45
  • $\begingroup$ Could the OP's statement be true for commuting matrices? (I guess so) $\endgroup$ Jan 23, 2015 at 18:00
  • $\begingroup$ @GottfriedHelms: Yes, for commuting matrices, the question will boil down to $|a^r - b^r| \le |a-b|^r$ for $a,b \ge 0$. $\endgroup$
    – Suvrit
    Jan 23, 2015 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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