# Hadamard product: Optimal bound on operator norm

Let $A,B$ be $n\times n$ matrices and denote by $A\star B$ the Hadamard product $(A\star B)(i,j)=A(i,j)B(i,j)$ (pointwise matrix multiplication). For $A$ positive definite it is known that $$\|A\star B\| \leq \sup_{i,j} |A(i,j)| \|B\|.$$ My question is what happens if we drop the positive definiteness assumption, i.e. what is the best constant $C>0$ such that $$\|A\star B\| \leq C \sup_{i,j} |A(i,j)| \|B\|$$ holds for arbitrary $n\times n$ matrices $A,B$. Is the constant $C$ independent of the size of the matrix $n$?

• Do you have a source or proof of the first inequality in the case where $A$ is positive definite?
– BenB
Jun 15, 2019 at 4:24
• For others who may be interested, the question of the inequality when $A$ is positive definite is addressed here math.stackexchange.com/q/3262944/119483
– ttb
Sep 16, 2022 at 16:57

The optimal such $$C$$ is $$\sqrt{n}$$, and so in particular there does not exist any such $$C$$ that is independent of $$n$$.
First, let me prove that $$C=\sqrt{n}$$ works. Let $$e_1,\dots,e_n$$ be the standard basis vectors, let $$v=\sum v_ie_i$$ be any vector, and let $$a=\sup_{i,j}|A(i,j)|$$. Note that for each $$i$$, $$\|(A\star B)e_i\|\leq a\|B\|$$, since $$\|Be_i\|\leq \|B\|$$ and $$(A\star B)e_i$$ is obtained from $$Be_i$$ by multiplying each entry by a scalar of size at most $$a$$. Thus $$\|(A\star B)v\|\leq\sum|v_i|\|(A\star B)e_i\|\leq a\|B\|\sum|v_i|.$$ By Cauchy-Schwarz, $$\sum|v_i|\leq \sqrt{n}\|v\|$$, so we conclude that $$\|A\star B\|\leq a\|B\|\sqrt{n}$$ and $$C=\sqrt{n}$$ works.
To prove $$C=\sqrt{n}$$ is optimal, let $$\omega$$ be a primitive $$n$$th root of unity and let $$B(i,j)=\frac{\omega^{ij}}{\sqrt{n}}$$. The columns of $$B$$ are orthonormal, so $$B$$ is unitary and $$\|B\|=1$$. Now let $$A(i,j)=\omega^{-ij}$$, so that $$A\star B$$ is the matrix whose entries are all $$\frac{1}{\sqrt{n}}$$. We have $$\|A\star B\|=\sqrt{n}$$ (the vector of all $$1$$s is an eigenvector of $$A\star B$$ with eigenvalue $$\sqrt{n}$$). Since $$|A(i,j)|=1$$ for all $$i,j$$, this means we must have $$C\geq \sqrt{n}$$.
• is there an example of the bound $\sqrt n$ with real matrices? Sep 15, 2021 at 9:15