# Showing a matrix is positive (semi) definite

I have a matrix, given by

$$\frac{1}{(1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2}}\begin{bmatrix} x_{1}^2 & x_{1}x_2 & x_{1}x_3 & \dots & x_{1}x_n \\ x_{2}x_1 & x_{2}^2 & x_{2}x_3 & \dots & x_{2}x_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n}x_1 & x_{n}x_2 & x_{n}x_3 & \dots & x_{n}^2 \end{bmatrix},$$ where $x_i \in \mathbb{R}$. I want to show that this is positive (semi)definite. There are many methods, naturally. I could show the determinant of the all the minors are positive, but how? I could try to work out the eigenvalues, but how? I could try and show this matrix is diagonally dominant with positive diagonals, but how? I could simply apply the definition of positive definite-ness, and see if the resulting number is non-negative, but it is a mess and it's not immediate to me why the result is non-negative.

I appreciate any help, thank you.

• Hint: the odd numbered methods you list are more practical than you think. – kimchi lover Feb 7 '18 at 19:42
• Which ones in particular? The minors determinant one is easy to show numerically but I’m not quite sure how I’d get the determinant easily by hand. Diagonally dominant I’m not sure will even work here (only for special cases on the choices of $x_i$. – Sorey Feb 7 '18 at 19:53
• It was a dumb comment. I meant, the even numbered ideas were easy, as in the answer below. – kimchi lover Feb 7 '18 at 20:27

$$A := \frac{1}{(1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2}}\begin{bmatrix} x_{1}^2 & x_{1}x_2 & x_{1}x_3 & \dots & x_{1}x_n \\ x_{2}x_1 & x_{2}^2 & x_{2}x_3 & \dots & x_{2}x_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n}x_1 & x_{n}x_2 & x_{n}x_3 & \dots & x_{n}^2 \end{bmatrix},$$
$$A(x) := \frac{1}{g(x)} xx^T, \quad g(x) := (1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2} > 0, \forall x \in \mathbb{R}^n$$ where $x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}$.
\begin{align*} z^T A(x) z &\geq 0, \forall z \in \mathbb{R}^n \Leftrightarrow \\ \frac{1}{g(x)} z^T x x^T z &\geq 0 \Leftrightarrow \\ \frac{1}{g(x)} (x^T z)^T (x^T z) = \frac{1}{g(x)} \| x^T z \|_2^2 &\geq 0 \end{align*} which is true for all $z \in \mathbb{R}^n$ by the nonnegativity of the norm and of the fact that $g(x) > 0, \forall x$.