TITLE: Nonnegative matrices and singular values
QUESTION [5 upvotes]: I would like to prove (or prove it is not true with a counter example) the following result:
Let $A$, $B$ be two squares matrices of size $n\times n$ with positive entries.
If $A \leq B$, then $\sum_{i = 1}^n \sigma_i(A) \leq \sum_{i = 1}^n \sigma_i(B)$. 
The sign $\leq$ between matrices is meant elementwise and the notation $\sigma_i(A)$ is used to denote the i-th singular value of $A$. The Perron-Frobenius theorem only provides information on the spectral radius, while the above statement involves all singular values.

REPLY [8 votes]: The inequality is not true, in general. Here is a counterexample:
Let $B =
  \begin{pmatrix}
    1 & 1 \\ 1 & 1
  \end{pmatrix}
$
and let $A = 
  \begin{pmatrix}
    1 & 1 \\ 0 & 1
  \end{pmatrix}
$ (my favourite counterexample matrix).
The singular values of $B$ are $2$ and $0$. A short computation shows that the singular values of $A$ are $(\frac{3 + \sqrt{5}}{2})^{1/2}$ and $(\frac{3 - \sqrt{5}}{2})^{1/2}$; the sum of those two values is strictly larger than $2$ (approximately $2.236$, but admittedly I used a calculator to check this...)