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In a recent post (Is this non-symmetric matrix positive definite?), I asked whether the following non-symmetric $n \times n$ matrix: $$ A = \begin{bmatrix} 1 & \delta_1 & \dots & \delta_1\\ \delta_2 & 1 & \dots & \delta_2\\ \vdots & \ddots & \ddots & \vdots\\ \delta_n &\dots & \delta_n& 1 \end{bmatrix}$$ where $\delta_i \in (0,1)$ for all $i$, was positive definite. The answer is no (provided by user151).

However, I realise that it would be sufficient for me if this matrix had all its principal minors positive. Which is obviously weaker. Does anyone have an idea of how to prove (or disprove) this?

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Let $D=\operatorname{diag}(\delta_1,\ldots,\delta_n)$. Then $A=Dee^T+I-D$ is a rank-one update of the matrix $I-D$. Hence $$ \det A =\det(I-D)\big(1+e^T(I-D)^{-1}De\big) =\det(I-D)\left(1+\sum_{i=1}^n\frac{\delta_i}{1-\delta_i}\right), $$ which is positive. Similarly, the other principal minors are positive too.

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  • $\begingroup$ Oh wow, fantastic ! Now I need to learn what a rank-one update is, and then why the formula follows from that observation. $\endgroup$
    – user20638
    Mar 30, 2023 at 15:05
  • $\begingroup$ Got it! Fantastic, thank you so much. $\endgroup$
    – user20638
    Mar 30, 2023 at 15:16

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