I have a real-valued symmetric square matrix of size $n$ with the following structure:
\begin{bmatrix}
(n-1)\alpha_{1}^2 & -\alpha_{1}\alpha_{2} & … & -\alpha_{1}\alpha_{n} \\
-\alpha_{1}\alpha_{2} & (n-1)\alpha_{2}^2 & … & -\alpha_{2}\alpha_{n} \\
\vdots & \vdots & \ddots & \vdots\\
-\alpha_{1}\alpha_{n} & -\alpha_{2}\alpha_{n} & … & (n-1)\alpha_{n}^2
\end{bmatrix}
I have the intuition that this matrix is always positive semi-definite. Empirically I am seeing that it is indeed the case but I am not sure how to prove it.
I have seen that the matrix will be positive semidefinite iff all the principal minors are non-negative. I am also seeing this empirically, but I am not able to show it formally. Any help is appreciated.
Best Answer
The matrix is equal to $D\left(nI-ee^T\right)D$, where $D=\operatorname{diag}(\alpha_1,\alpha_2,\ldots,\alpha_n)$ and $e=(1,1,\ldots,1)^T$. Hence it is positive semidefinite, because by Cauchy-Schwarz inequality, $$ x^TD\left(nI-ee^T\right)Dx =n\|Dx\|^2-\langle Dx,e\rangle^2 =\|e\|^2\|Dx\|^2-\langle Dx,e\rangle^2\ge0. $$