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](https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem) only provides information on the spectral radius, while the above statement involves all singular values.



Nonnegative matrices and singular values
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...)


