The hadamard-product tag has no usage guidance.

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### Billera Tree Space

I am studying the tree space of Billera and I do not really understand why it is an Hadamard Space. I have already read L. Billera, S. Holmes, K. Vogtmann, Geometry of the space of phylogenetic trees, ...

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### Hadamard Product and Eigendecomposition

I just found this related question in here Q1.
Given a positive definite matrix $\mathbf{A}$, consider its eigendecomposition $(\mathbf{A}\mathbf{V} = \mathbf{V}\mathbf{D})$. Consider an arbitrary ...

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### Prime Hadamard Matrices

Assume that $n$ is a sufficiently large number. Is there a Hadamaed matrix $H_{4n \times 4n}=(h_{ij})$ with the last row and the last cloumn $J$ (thet is, for every $k$, $h_{k,4n}=1$ and $h_{4n, ...

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### A Hadamard product of the zeros of the Riemann integral. Does it put any constraints on where the $\rho$'s can reside in the critical strip?

I have deleted a previous, now obsolete question on the same topic.
Take the well-known Riemann integral:
$$\displaystyle \pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) ...

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### How to solve a matrix equation with both inverses and a hadamard product?

I have a matrix equation of the form:
$$
A^{-1} = B + A \circ C
$$
where $\circ$ denotes the Hadamard product (i.e., $(A\circ C)_{ij} = A_{ij}B_{ij}$). How can I determine if a solution for $A$ ...

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### Inequalities for Hadamard products of complex symmetric matrices

Consider a complex symmetric matrix $$ C= C_R + i C_I $$ with $C_R,C_I \in \text{Mat}_{n\times n}(\mathbb R)$ symmetric, and assume that the eigenvalues of $C_R$ are all strictly positive. Then, $C$ ...

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### Hadamard product and inertia

One of Schur's famous results says that if $A,B$ are positive semidefinite matrices, then the Hadamard (i.e. entrywise) product $A \circ B$ is also positive semidefinite. It's also true if "semi" is ...

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### Generalizations of Oppenheim's inequality

The well-known Oppenheim inequality says that for two positive definite matrices $A,B$ it holds that $\det(A \circ B) \geq (\prod{a_{ii}})\det(B)$.
There has been a lot of beautiful work done ...