Subadditivity of the square root for matrices

Question

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 $0<r<1$.

Answer 1

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.