For what values of $a \in \mathbb{R}$ is this quadratic form positive semidefinite?

Question

Let the quadratic form $f \in \mathbb{R}[x_1, \dots, x_n]$ be defined by

$$ f(x_1, \dots, x_n) := \sum_{i=1}^{n}{x_i^2} + a \sum_{i=1}^{n} \sum_{j=i+1}^{n} {x_i x_j}$$

where $a>0$. I want to know for which values of $a$ it is verified that $f(x_1, \dots, x_n)\ge 0$ for all $(x_1, \dots, x_n) \in \mathbb{R}^n$. And I would like to have a formal proof of it.

I am thinking about considering $ f (\textbf{x}) = {\bf x}^\top {\bf A} \, {\bf x} $, where

$${ \bf A} = \begin{bmatrix} 1 & \frac{a}{2} & \frac{a}{2} & \cdots & \frac{a}{2} \\ \frac{a}{2} & 1 & \frac{a}{2} & \cdots & \frac{a}{2} \\ \frac{a}{2} & \frac{a}{2} & 1 & \cdots & \frac{a}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{a}{2} & \frac{a}{2} & \frac{a}{2} & \cdots & 1 \end{bmatrix} $$

I get that the eigenvalues are $\lambda_1 = 1 + \frac{a}2(n-1)$ and $\lambda_2 = \dots =\lambda_n = 1 - \frac{a}2$. I see that only if $a>2$ I have negative eigenvalues, but I am not sure how does that relate with the sign of the quadratic form. With some graphical and numerical tests, I think for $a < 2$ I have $f > 0$, while for $a > 2$ it is possible to have negative results. Could anyone help me to have a formal proof?

Answer 1

Let

$$ 2 \, {\bf A} (\alpha) := \alpha \, {\bf 1}_n {\bf 1}_n^\top + (2 - \alpha) \, {\bf I}_n = n \alpha \left( \frac{{\bf 1}_n}{\sqrt{n}} \right) \left( \frac{{\bf 1}_n}{\sqrt{n}} \right)^\top + (2 - \alpha) \, {\bf I}_n $$

We are interested in finding the ($1$-dimensional) spectrahedron $\left\{ \alpha \in {\Bbb R} \mid {\bf A} (\alpha) \succeq {\bf O}_n \right\}$. Note that, in the definition of $2 \, {\bf A} (\alpha)$, the first term is the single-term spectral decomposition of a symmetric rank-$1$ matrix. Let $\bf Q$ be a $n \times n$ orthogonal matrix whose first column is the unit vector $\frac{1}{\sqrt{n}} {\bf 1}_n$. Thus,

$$ \begin{aligned} 2 \, {\bf A} (\alpha) &:= {\bf Q} \left( \textbf{diag} \left( n \alpha, 0, \dots, 0 \right) + (2 - \alpha) {\bf I}_n \right) {\bf Q}^\top \\ &\,\,= {\bf Q} \, \textbf{diag} \left( n \alpha + (2 - \alpha), 2 - \alpha, \dots, 2 - \alpha \right) {\bf Q}^\top \end{aligned} $$

Hence,

$$ \left\{ \alpha \in {\Bbb R} \mid {\bf A} (\alpha) \succeq {\bf O}_n \right\} = \left\{ \alpha \in {\Bbb R} \mid \left( n \alpha + 2 - \alpha \geq 0 \right) \land \left( 2 - \alpha \geq 0 \right) \right\} = \color{blue}{\left[ -\frac{2}{n-1}, 2 \right]} $$

Answer 2

No need of matrix here.

I assume you want $f(x_1,\dots,x_n)\ge 0$, otherwise the strict inequality is false on zero vector, as pointed out by Tzimmo. $$f(x_1,\dots,x_n)=\left(1-\frac a2\right)\sum_{i=1}^nx_i^2+\frac a2\sum_{i,j}x_ix_j\\ =\left(1-\frac a2\right)nV_x+\left(\frac{an^2}{2}+n-\frac{an}{2}\right)\overline x^2$$

Then as soon as the mean is 0, then you need to have $a\le 2$.

Conversely, for all $a\le 2$ the equality above shows that $f$ is nonnegative.