TITLE: Subadditivity of the square root for matrices
QUESTION [6 upvotes]: 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 $0<r<1$.

REPLY [4 votes]: 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.