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?
2 Answers
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.
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 $\langle x, \xi \rangle = 0$.
