Is this non-symmetric matrix positive definite?

Question

The Linear Complementarity Problem $\mbox{LCP} (q,A)$ has nice properties when the matrix $A$ is positive definite, even if $A$ is not symmetric. I am concerned with an LCP and would like to know if my non-symmetric matrix is positive definite.

Assume $J$ is an $n \times n$ matrix of ones, and $\Delta = \mbox{diag} (\delta_1, \dots, \delta_n)$, where $\delta_i \in (0,1)$ for all $i$. Let $$ A := I + \Delta (J-I) = \begin{bmatrix} 1 & \delta_1 & \dots & \delta_1\\ \delta_2 & 1 & \dots & \delta_2\\ \vdots & \ddots & \ddots & \vdots\\ \delta_n &\dots & \delta_n& 1 \end{bmatrix}$$

I think this matrix is positive definite (i.e., $x^T A x > 0$ for all $x\neq 0$), but I cant' prove it. Any idea?

Answer 1

No, your $A$ does not always possess a positive definite symmetric part. Consider $B=\pmatrix{I_{n-1}&0\\ e^T&1}$ first, where $e\in\mathbb R^{n-1}$ denotes the vector of ones. When $n\ge6$, we have $$ \det\frac{B+B^T}{2} =\det\pmatrix{I_{n-1}&\frac12e\\ \frac12e^T&1} =1-\frac14e^Te=1-\frac{n-1}{4}<0. $$ Hence $B$ has not a positive semidefinite symmetric part. In turn, every matrix $A$ close to $B$ also does not possess any positive definite symmetric part.

Answer 2

Your definition of positive definiteness of $A$ is equivalent to the usual one applied to the symmetric matrix $\frac{1}{2}(A+A^T).$ Clearly your $A$ cannot be positive definite if for instance $\frac{1}{2}(\delta_1+\delta_2)>1.$