# Pointwise product of real symmetric nonnegative matrices

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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I don't have the time to write an answer right now, but you should try to show that $S_n^{+}$ (ie. symmetric positive semi-definite matrices) is stable by this product (call it $\otimes$), one way to see that is to decompose the two matrices in their spectral projectors (with the eigenvalues as coefficients) and show that two orthogonal projections behave well under $\otimes$. – Najib Idrissi Feb 15 '12 at 16:29
Here's the reference given on the Wikipedia page: buzzard.ups.edu/courses/2007spring/projects/million-paper.pdf – Jonas Meyer Feb 15 '12 at 19:16

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.
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 $A^{1/2}BA^{1/2}$.