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Let $A,B$ be complex matrices of the same size. I am looking for some references on the comparison of $|A\circ B|$ and $|A|\circ| B|$, where $|A|=(A^*A)^{1/2}$, "$\circ$" stands for Hadamard product.

The comparison include eigenvalue majorization, norm and....

Thank you.

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    I checked that, but there is no comparison between $|A\circ B|$ and $|A|\circ|B|$.2011-07-11

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@Sunni: the inequality $x^*|A\circ B|x\le x^*(|A|\circ |B|)x$ can fail.

Here's an example. Let $A=\begin{bmatrix}1&t\\0&0\end{bmatrix}$ where $t$ is a large real number, let $B=I_2$ and let $x= \begin{bmatrix}1\\0\end{bmatrix}$. Then $A\circ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ so $x^* |A\circ B|x=1$, but $|A|=\begin{bmatrix}1&t\\t&t^2\end{bmatrix}^{1/2}=t \begin{bmatrix} t^{-2}&t^{-1}\\t^{-1}&1 \end{bmatrix}^{1/2}\approx t \begin{bmatrix} 0&0\\0&1 \end{bmatrix}$, so $|A|\circ |B|\approx t \begin{bmatrix} 0&0\\0&1 \end{bmatrix}$ and $x^*(|A|\circ |B|)x \to 0$ as $t\to \infty$.

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    @Sunni: sorry, I don't know any references for this sort of question. But the eigenvalue comparison doesn't seem to work out: if you take $t=1$ above, then $1/\sqrt2$ is the only eigenvalue of $|A|\circ |B|$ (with multiplicity $2$), but the eigenvalues of $|A\circ B|$ are $0$ and $1$.2011-07-12
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The comparison you want applies to any kind of tuples of numbers, whether vectors or arranged in blocks like a matrix. Namely, the Cauchy-Schwarz-Bunyakowsky inequality is $ \Big(\sum_n |a_nb_n|\Big)^2 \le \sum_n |a_n|^2\;\cdot \sum_n |b_n|^2 $

[Edit: additions] There are obviously many questions one could ask about the particulars... For example, for set-up, for $A$ self-adjoint, $A=UDU^{-1}$ with $U$ unitary and $D$ real diagonal, and the "absolute value" $|A|$ is $U|D|U^{-1}$ where $|D|$ has absolute values of $D$'s entries. Ok, and then the trace-norm of $A$ is the sum of the absolute values of eigenvalues, and the Hilbert-Schmidt norm of $A$ is the square root of the sum of squares of absolute values of eigenvalues. When $A,B$ are simultaneously diagonalizable (e.g., commute and are self-adjoint), then the Hadamard product is just the product, and $|A\circ B|=|A|\circ |B|$. In this simple case, Cauchy-Schwarz-B. implies that the HS-norm of $AB$ is at most the product of the HS norms. (This was the content of my earlier.) When $A,B$ do not necessarily commute, but at least one of them is self-adjoint (or normal), we can still take one (say $A$) to be diagonal, without loss of generality, and then the Hadamard product $A\circ B$ kills off all but the diagonal entries of $B$. Killing off entries reduces norms, etc.

Is this a relevant amplification?

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    The examples you considered were too special..... Perhaps I need to state my problem more specific. Is it true $x^*|A\circ B|x\le x^*(|A|\circ |B|)x$ for all column vector $x$? $A,B$ are arbitrary complex matrices of the same size.2011-07-12