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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?

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  • $\begingroup$ Are you sure that it makes sense to speak of positive definiteness for non-symmetric matrices? $\endgroup$ Commented Mar 29, 2023 at 6:41
  • $\begingroup$ Related $\endgroup$ Commented Mar 29, 2023 at 6:41
  • $\begingroup$ 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" $\endgroup$ Commented Mar 29, 2023 at 6:42
  • $\begingroup$ What is the motivation for this question? $\endgroup$ Commented Mar 29, 2023 at 7:58
  • $\begingroup$ 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. $\endgroup$
    – user20638
    Commented Mar 29, 2023 at 8:17

2 Answers 2

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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 $\frac{1}{2}(\delta_1+\delta_2)>1.$

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  • $\begingroup$ Thank you. However since both $\delta_1$ and $`\delta_2$ are smaller than 1 this will not happen $\endgroup$
    – user20638
    Commented Mar 29, 2023 at 7:56
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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – user20638
    Commented Mar 29, 2023 at 7:58
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Commented Mar 29, 2023 at 9:31
  • $\begingroup$ Yes, thank you. Unfortunately, most of the time I will have that the $\delta$'s are "large", possibly all close to 1. $\endgroup$
    – user20638
    Commented Mar 29, 2023 at 10:40

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