Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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: – 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.

share|cite|improve this answer

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".

share|cite|improve this answer
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). – Jonas Meyer Feb 15 '12 at 19:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.