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In an article the author consider the square matrix $(A^{ij}(x))_{i,j=1}^d$ defined by $$ A^{ij}(x):= \frac{x_ix_j}{|x|^2} \qquad (x\in R^d). $$ He writes that the matrix is elliptic. I don't know to prove that there exists $c>0$ such that $$ \sum_{i,j=1}^d A^{ij}(x)\xi_i\xi_j\ge c|\xi|^2 \qquad \forall x,\xi\in R^d. $$ Someone can help me?

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    $\begingroup$ I saw something similar in the counterexamples to the solution to the 19th and 20th Hilbert problems given by De Giorgi, Giusti and Miranda and Maz'ya, but there are other terms in such counterexamples: could you give us the complete paper reference? $\endgroup$ May 1, 2019 at 16:25
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    $\begingroup$ You are right. I'm reading these type of papers ( for example "An irregular complex valued solution to a scalar unifomly elliptic equation", author is Frehse). I did not consider the constant terms, and so i was wrong. Now it is clear. Thank you all. $\endgroup$
    – Revzora
    May 2, 2019 at 8:37

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Τhe inequality, that you wrote down, is not true in general. First of all, it is equivalent to

$$\langle x,\xi\rangle^2=\left(\sum_{i=1}^dx_i{\xi}_i\right)^2=\sum_{i,j=1}^d(x_i{\xi}_i)(x_j{\xi}_j)\geq c|x|^2|\xi|^2,\ \ \forall x,\xi\in \mathbb{R}^d.$$

If this is true for some $c>0$, we can take two non-zero vectors $x$ and $\xi$ that are perpendicular (like $x=(1,0,...,0)$ and $\xi=(0,1,0,...,0)$) and in this case the usual inner product of $\mathbb{R}^d$ at the leftmost side is $0$. Then, it follows that $0\geq c|x|^2|\xi|^2>0,$ which is a contradiction.

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For any $x$, the matrix is positive semidefinite, but not positive definite. This is because $$ \sum_{i,j} A_{ij}(x)\xi_i \xi_j = \frac{|\langle x, \xi \rangle|^2}{|x|^2} \ge 0 $$ with equality if $\langle x, \xi \rangle = 0$.

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