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!