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I have encoutered several times the following claim :

Let $A$ and $B$ be two real symmetric matrices (of dimension $n\times n$) with nonnegative coefficients and such that their eigenvalues are nonnegative. Let $C$ be their pointwise product, i.e. the $n\times n$ matrix with coefficients $C_{ij}=A_{ij}B_{ij}$. Then the matrix $C$ has nonnegative eigenvalues.

It seems to be true, but I can't find a proof. Does anybody know how to show this ?

Also, this result implies that if we denote by $A^{[n]}$ the matrix with coefficients $A^{[n]}_{ij}=(A_{ij})^n$ with $n\in \mathbb{N}$, then $A^{[n]}$ has nonnegative eigenvalues. Does this result still hold if we only suppose that $n\in\mathbb{R}^+$ ? If it doesn't hold anymore, is there a sufficient condition weaker than $n\in\mathbb{N}$ under which the result holds ?

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    Here's the reference given on [the Wikipedia page](http://en.wikipedia.org/wiki/Hadamard_product_%28matrices%29): http://buzzard.ups.edu/courses/2007spring/projects/million-paper.pdf2012-02-15

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You only need $A, B$ be non-negative (entry-wise) in order to make sure that $C$ is non-negative. But it's not needed to prove the positive semi-definiteness. Here is the idea from Horn and Jonson's book titled: Matrix Analysis:

Let $\otimes $ denote the entry-wise product. If $A_{n\times n}$ is a PSD matrix of rank $k$, then it can be written as $A=v_1v_1^*+\dots + v_kv_k^*.$ Similarly $B=w_1w_1^*+\dots + w_mw_m^*.$ Let $u_{ij}=v_{i} \otimes w_{j}$ Then $A \otimes B = \sum_{i,j=1}^{k,m} u_{ij}u_{ij}^*$ which is a sum of rank 1 PSD matrices. So, it is PSD.

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Your $A, B$ are actually positive semidefinite, thus the eigenvalues of $AB$ are nonnegative. Since the eigenvalues of $AB$ is the same as those of $A^{1/2}BA^{1/2}$.

Not sure what you mean by "nonnegative coefficients".

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    Anonymous is not asking about the ordinary matrix product, but rather the entrywise product, a.k.a. Schur product or Hadamard product. By "coefficients" I think Anonymous means "entries" (but the hypothesis that the entries are nonnegative is unnecessary for the first question).2012-02-15