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For two Matrices $A,B \in \mathbb{R}^{m \times n}$ the Hadamard Product is defined as $(A \circ B)_{i,j} = A_{i,j}B_{i,j}$.

For a proof of convergene I require an upper (and ideally a lower) bound on \begin{align} \sum_{i>k}^{\min(m,n)}{\sigma_i^2(A \circ B)} \end{align} for Matrices with $|A_{i,j}| < 1$ and $|B_{i,j}| < 1$ for all $i,j$ and an arbitrary $1 \leq k < \min(m,n)$

From tests with randomly generated matrices I suspect that the following strong pointwise property, that could be used for a good upper bound, holds:

\begin{align} \sigma_i(A \circ B) < \text{max}(\sigma_i(A),\sigma_i(B)) \end{align}

Generally a decrease in the singular values is expected due to the well known fact that $\lVert A \rVert_F = \sqrt{\sum_i{\sigma_i^2(A)}}$ and by definition $\lVert A \circ B \rVert_F < \min(\lVert A \rVert_F,\lVert B \rVert_F)$, but this does not imply either of the inequalities stated above.

Is there some known upper bound I could use for this problem? Does someone see a proof or counterargument to the strong bound I provided?

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    $\begingroup$ Obviously false even for the norm: take $A=B=\begin{bmatrix}1&1\\-1&1\end{bmatrix}$. Computers spoiled us entirely: we run hundreds of random tests on 10 by 10 instead of thinking for 5 minutes about 2 by 2 :-( $\endgroup$
    – fedja
    Commented Jun 5, 2018 at 14:07
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    $\begingroup$ It seems that the inequalities in your question (both for the singular values and for the Frobenius norm) do not scale correctly: If $A = B = t I$ (where $I$ is the $n\times n$-identity matrix and $t \in [0,\infty)$), then the singular values and the Frobenius norm of $A$ and $B$ grow linearly in $t$, while the singular values and the Frobenius norm of $A \circ B$ grow quadratically in $t$. Maybe you forgot a square root somewhere? $\endgroup$
    – Jochen Glueck
    Commented Jun 5, 2018 at 15:13
  • $\begingroup$ As i presumed $|A_{i,j}|,|B_{i,j}|<1$ this does not matter since for $t<1$ obviously $t^2<t$ holds. @fedja is right that this does not work for their example, though I wonder if this only happens for some null subset of $\mathbb{R}^{m \times n}$. I would still be interested in a good upper bound for the first sum. $\endgroup$
    – Jabor
    Commented Jun 5, 2018 at 15:41
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    $\begingroup$ I would still be interested in a good upper bound for the first sum. Define "good". In other words, state precisely what quantities you know, in terms of which the estimate should be given. $\endgroup$
    – fedja
    Commented Jun 5, 2018 at 15:47
  • $\begingroup$ I see. Apparently, I overlooked your condition on the entries of $A$ and $B$ (still, the scaling behaviour of the inequalities seems strange to me, but this might be due to the fact that I don't have much experience with Hadamard products). Considering your question on whether @fedja's counterexample only occurs on a nullset: By a perturbation argument, the same problem occurs on an open neighbourhood of fedja's matrices (and this neighbourhood intersects the set of matrices which fulfil your condition on the entries in a nonempty open set). $\endgroup$
    – Jochen Glueck
    Commented Jun 5, 2018 at 15:56

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Doubtlessly, reading the article will shed more light on your question.

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    $\begingroup$ The result of this paper by X. Zhan is stronger than the mentioned paper. $\endgroup$
    – Mahdi - Free Palestine
    Commented Jun 5, 2018 at 18:04
  • $\begingroup$ Why embed a screenshot of the paper rather than simply linking to it? $\endgroup$
    – j.c.
    Commented Jun 5, 2018 at 19:08
  • $\begingroup$ @j.c. What, precisely, is your objection? $\endgroup$
    – Igor Rivin
    Commented Jun 5, 2018 at 20:09

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