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Is there a known relation/formula for $$(A\circ B, C)$$ where $\circ$ is the Hadamard product and $(\cdot, \cdot)$ is the scalar (euclidean) product?

In particular, I have a vector $y$ and a two dimensional subspace $V$ of $R^n$ such that $(v,y)=0$ for any $v \in V$.

What properties do I need on $y$ to ensure that there exists an $z$ in $V$ such that $$(z,y)=0 \quad \text{and}\quad (z\circ z,y)\geq 0$$

Is there a known formula/result which can help in this direction?

NB. A trivial solution is $y= \text{componentwise nonnegative}$. Can we assume any weaker property on $y$?

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The conjecture seems to be false as stated. Take e.g. $n = 3$, y = $\[0,0,1\]$ and V hence the X/Y plane. Then for any $z \in V $ also $z \circ z \in V$ and hence $(z \circ z, y) = 0$. So in this case there exists no $z$ with the given properties. – veryltdbeard Mar 18 '13 at 13:11

1 Answer 1

I can't recall any such formula, but your claim is always true if you pick $z=0$, and in general false if $z\neq0$. For a counterexample, consider $V=\operatorname{span}\{u=(2,1,0,0)^T,\,v=(0,0,2,1)^T\}$ and $y=(-1,2,-1,2)^T$. Now for any $z=au+bv$ with $(a,b)\neq(0,0)$, $\langle z\circ z,y\rangle=-2(a^2+b^2)<0$.

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ok, thanks. I need some further hypothesis on y. I am going to edit the question above. – F. T. Mar 18 '13 at 16:30

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