how to show that $A=[a_i+a_j]_{ij}$ has exactly one positive and one negative eigenvalue.

Question

Let $a_1,\dotsb,a_n$ are not all equal positive real numbers such that $A=[a_i+a_j]_{ij}^n$ is a real square matrix, i.e. $$A=\left[\begin{matrix}a_1+a_1 &a_1+a_2&\dotsb &a_1+a_n \\ a_2+a_1 &a_2+a_2 &\dotsb &\vdots \\ \vdots & \cdots \\a_n+a_1&a_n+a_2 &\cdots & a_n+a_n \end{matrix}\right].$$ How to show that A has exactly one positive and one negative eigenvalue.

Answer 1

Denote $$ \alpha = [a_1, a_2, \cdots, a_n]^T, \beta = [1, 1, \cdots, 1]^T$$ so $ A = \alpha\beta^T + \beta\alpha^T$ is a rank-2 matrix, given $\alpha\neq\beta$. $Ax = 0$ has non-zero solution so 0 is the eigen value of it and its geometric multiplicity is $n-2$.

Then we can check $||\beta||\alpha + ||\alpha||\beta$ and $||\beta||\alpha - ||\alpha||\beta$ are eigen vectors too and the corresponding eigen value are $\lambda_1 = ||\alpha||||\beta|| + \alpha^T\beta$ and $\lambda_2=-||\alpha||||\beta|| + \alpha^T\beta$. By cauchy inequality: $$ \left<\alpha, \beta\right> = \alpha^T\beta\leq ||\alpha||\cdot||\beta||$$ we can see $\lambda_1 >0 $ and $\lambda_2 < 0$.