TITLE: Majorization relation between the absolute values squared of the entries of a matrix and the singular values squared
QUESTION [7 upvotes]: Let $A = [A_{ij}]$ be an $n\times n$ square matrix with complex entries, and let $\sigma_k$, $k=1,\ldots, n$ be its singular values. Suppose that the squared Frobenius norm satisfies 
$$
\mathrm{Tr}(A^\dagger A) = \sum_{i,j=1}^{n}|A_{ij}|^2 = \sum_{k=1}^n\sigma^2_k=1 \>,
$$
where $A^\dagger$ is the conjugate transpose of $A$. 
Is the vector given by the absolute values squared of the entries, $(|A_{ij}|^2)_{ij}$, majorized by the vector $(\sigma_k^2)_k$? (As usual when discussing majorization, with proper padding of 0's, so that both vectors have $n^2$ elements.)
Consider vectors $(a_i)$ and $(b_i)$ of length $m$ such that $\sum_{i=1}^m a_i = \sum_{i=1}^m b_i$. Then, we say that $a$ majorizes $b$ if $\sum_{i=1}^k a_{(i)} \geq \sum_{i=1}^k b_{(i)}$ for each $1 \leq k \leq m$, where $a_{(i)}$ denotes the $i$th largest element of $(a_i)$ and likewise for $b_{(i)}$ with respect to $(b_i)$.
For a more detailed definition of majorization, please see http://en.wikipedia.org/wiki/Majorization . 
I looked numerically for a counterexample, and found none. If it is true, I would suppose it is well-known, and in case I would appreciate a reference as precise as possible.
Thank you!

REPLY [2 votes]: NOTE: Let us see if I can post this answer now :) 

First, let me note that the condition that the Frobenius norm is one is irrelevant, since multiplying the matrix by some positive constant changes both the entries and the singular values by the same multiplicative constant. Anyway, for the same reason, we can restrict to the case mentioned in the question.
The solution is more easily seen in the language of quantum information. The entries $A_{ij}$ of the matrix can be seen as the coefficients of a bipartite state vector $|\psi\rangle=\sum_{ij}A_{ij}|i\rangle|j\rangle$, where $\{|i\rangle\}$ and $\{|j\rangle\}$ are (local) orthonormal bases. Suppose that the largest $k$ entries $|A_{ij}|^2$ correspond to pairs of indexes $(i,j)\in I_k$, where $I_k$ is some $k$-element subset of the set $\{ (i,j)|i,j=1,\ldots,n \}$. Then
$$
\sum_{(i,j)\in I_k}|A_{ij}|^2 = \sum_{(i,j)\in I_k}\textrm{Tr}(|i\rangle\langle i|\otimes |j\rangle\langle j| |\psi\rangle\langle\psi|)
$$
This quantity is less than
$$
\sum_{i:\exists j s.t. (i,j)\in I_k}\textrm{Tr}(|i\rangle\langle i|\otimes I |\psi\rangle\langle\psi|)=\textrm{Tr}_1(\sum_{i:\exists j s.t. (i,j)\in I_k}|i\rangle\langle i|\textrm{Tr}_2(|\psi\rangle\langle\psi|))
$$
where $I$ denotes the identity matrix/operator and we have split the trace into the trace onto the first system and onto the second system. The matrix $\textrm{Tr}_2(|\psi\rangle\langle\psi|)$ is positive semidefinite, with eigenvalues corresponding to the squares of the singular values of $A$ (it is the reduced state of $|\psi\rangle$ in the language of quantum information), and we have
$$
\textrm{Tr}_1(\sum_{i:\exists j s.t. (i,j)\in I_k}|i\rangle\langle i|\textrm{Tr}_2(|\psi\rangle\langle\psi|))\leq \max_{P_k} \textrm{Tr}_1(P_k\textrm{Tr}_2(|\psi\rangle\langle\psi|))=\sum_{i=1}^k\sigma_k^2,
$$
where the maximum is over projections of rank $k$. Since this is valid for any $k$, the conjecture is proven.