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!
We will actually prove something stronger and then see that the desired result follows.
Let $A$ be defined as in the problem and take $\newcommand{\Atr}{A^{\dagger}}B = \Atr A$. Let $a_{ij}$ denote the $(i,j)$th element of $A$. Note that the diagonal elements of $B$, which we denote by $b_{ii}$ are real nonnegative numbers. We assume without loss of generality that $b_{ii} \geq b_{i+1,i+1} \geq 0$ for all $1 \leq i < n$.
Claim 1: The diagonal of $B$ majorizes $(|a_{ij}|^2)_{ij}$.
Let $s_k = \sum_{i=1}^k b_{ii}$ denote the sum of the $k$ largest $b_{ii}$. For each $i$, $b_{ii} = \sum_{j=1}^n |a_{ji}|^2$ is the sum of the squared moduli of the elements in the $i$th column of $A$. Consider the $k$ largest $|a_{ij}|^2$. Then, these $k$ elements lie within no more than $k$ unique columns of $A$ and, so, the sum of these $k$ elements is clearly less than or equal to the sum of the corresponding $b_{ii}$'s. But, this latter sum is definitely smaller than $s_k$. This holds for each $1 \leq k \leq n$ and so the claim is established since also $\sum_{i=1}^n b_{ii} = \sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^2$.
Theorem (von Neumann): Let $S$ and $T$ be arbitrary complex-valued $n \times n$ matrices. Let $(\sigma_i)$ and $(\tau_i)$ be the singular values of $S$ and $T$, respectively, in nonincreasing order. Then, $|\mathrm{Tr}(ST)| \leq \sum_{i=1}^n \sigma_i \tau_i$.
A nice, elementary proof of this can be found in
We don't actually need such a strong statement to prove the next claim, but I give the result above because it's very nice and doesn't seem to be as well-known as it should be.
Claim 2: Let $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \geq 0$ be the eigenvalues of $B$. Then $(\lambda_i)$ majorizes $(b_{ii})$.
$B$ has an eigendecomposition such that $B = Q \Lambda Q^*$ where $Q$ is unitary and $\Lambda$ is a diagonal matrix corresponding to $(\lambda_i)$. Now, note that $$\newcommand{\Tr}{\mathrm{Tr}} s_k = \Tr(J_k^T B J_k) $$ where $$\newcommand{\zmat}{\mathbf{0}} J_k = \left(\begin{matrix} I_k \\ \zmat \end{matrix}\right) $$ with $I_k$ being a $k \times k$ identity matrix and $\zmat$ being an $(n-k) \times k$ all-zeros matrix.
Then $$ \Tr(J_k^T B J_k) = \Tr(J_k^T Q \Lambda Q^* J_k) = \Tr(Q^* J_k J_k^T Q \Lambda) \>. $$
Observe that $Q^* J_k J_k^T Q$ is a Hermitian matrix with $k$ singular values that are one and $n-k$ that are 0, and so by von Neumann's theorem, we get that $$ s_k = |s_k| = |\Tr(J_k^T B J_k)| \leq \sum_{i=1}^k 1 \cdot \lambda_i = \sum_{i=1}^k \lambda_i \> . $$
Since this holds for each $k \leq n$, the claim is established.
Epilogue: Combining Claims 1 and 2 gives the desired result as stated in the question. But, Claim 2 by itself is actually quite a bit stronger. Also, as alluded to above, more elementary means can be used to show Claim 2 and they are almost as easy as wielding von Neumann's somewhat bigger hammer.