# Is this non-symmetric matrix positive definite?

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?

• Are you sure that it makes sense to speak of positive definiteness for non-symmetric matrices? Commented Mar 29, 2023 at 6:41
• Related Commented Mar 29, 2023 at 6:41
• Quoting Qiaochu Yuan: "positive-definite should not be a term that applies to matrices. It should only apply to quadratic forms, which are naturally described by symmetric matrices only" Commented Mar 29, 2023 at 6:42
• What is the motivation for this question? Commented Mar 29, 2023 at 7:58
• There are results on LCP (Linear Complementary Problem) that I would like to use. They apply to non-symmetric positive definite matrices or, if you don't like the term, to matrices such that $x^T A x > 0$ for all $x \neq 0$. Unfortunately @user1551 answered my question by the negative. Commented Mar 29, 2023 at 8:17

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 $$\frac{1}{2}(\delta_1+\delta_2)>1.$$
• Thank you. However since both $\delta_1$ and $`\delta_2$ are smaller than 1 this will not happen Commented Mar 29, 2023 at 7:56
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.
• Thank you. And of course $x^T \frac{B+B^T}{2} x < 0$ implies $x^T A x < 0$. Thank you for this counter-example. Any idea of conditions on the $\delta$'s such that this matrix would be positive definite ? Commented Mar 29, 2023 at 7:58
• @user20638 The most obvious sufficient condition is $\delta_i<\frac{1}{n-1}$ for all $i$, so that $A+A^T$ is strictly diagonally dominant. If one employs a generalised version of Gershgorin disc theorem, the requirement in this sufficient condition may probably be weakened. Commented Mar 29, 2023 at 9:31
• Yes, thank you. Unfortunately, most of the time I will have that the $\delta$'s are "large", possibly all close to 1. Commented Mar 29, 2023 at 10:40