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Given a matrix \begin{equation} M = \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix} \end{equation} and $a,b,c$ are not all equal and are all positive. We HAVE That \begin{equation} Det(M)=a^3+b^3+c^3-3abc>0 \end{equation} where the inquality is true by AM-GM inequality.

Now if we perturb the matrix a little bit, say \begin{equation} M_1 = \begin{bmatrix} a_1 & b_1 & c_1 \\ c_2 & a_2 & b_2 \\ b_3 & c_3 & a_3 \end{bmatrix} \end{equation} where the relative order remains unchanged, say $\forall i, j, k\in\{1,2,3\}$, $0<a_i<b_j<c_k$ if previously $a<b<c$, I want to show a similar positive determinant \begin{equation} Det(M_1)=a_1a_2a_3+b_1b_2b_3+c_1c_2c_3-a_1b_2c_3-a_2b_3c_1-a_3b_1c_2>0 \end{equation}

YYF
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  • very good question. I suggest you start with a rank 1 update, where the variation is linear : https://math.stackexchange.com/questions/219731/determinant-of-rank-one-perturbations-of-invertible-matrices – G. Fougeron Jul 06 '21 at 13:27
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    Don't you need $a,b,c$ to be positive? Otherwise $a=-2, b=c=-1$ is a counterexample to your first inequality. – Zerox Jul 06 '21 at 13:42
  • Thank you for pointing out! I forgot this condition. – YYF Jul 06 '21 at 13:53
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    @YYF I know an inequality could immediate solve this problem, sadly I don't find any English reference on it. If you have a translator you can read this. – Zerox Jul 06 '21 at 14:09
  • @Zerox Thank you for the reference. It seems to be called as rearrangement inequality in English https://en.wikipedia.org/wiki/Rearrangement_inequality . – YYF Jul 07 '21 at 07:23

1 Answers1

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Thanks to the reference given by @Zerox, the porblem can be solve using the inequality mentioned above.

However, for the strict inequality, a sufficient condition may be given by $a_i\le b_j\le c_k$, and either $\max a_i< \min b_j$ or $\max b_j<\max c_k$ is satisfied.

A two dimensional case can be shown using $(b_j-a_j)(b_i-a_i)>0$, and rearranging terms, $$a_ia_j+b_ib_j>a_ib_j-a_jb_i$$ now \begin{align} a_1a_2a_3+b_1b_2b_3+c_1c_2c_3&>a_1b_2b_3+a_2a_3b_1+c_1c_2c_3\\ &>a_1b_2c_3+a_2a_3b_1+c_1c_2b_3\\ &>a_1b_2c_3+a_3b_1c_2+a_2b_3c_1 \end{align} all inequality holds by the two dimensional case, the first exchange first two terms using $a_1<b_1$ and $a_2a_3<b_2b_3$.

The second inequality exchange first and third terms, using $a_1b_2<c_1c_2$ and $b_3<c_3$.

And the last inequality use $a_3b_1<b_3c_1$ and $a_2<c_2$.

If one of the above inequality is a strict larger one, the strictly greater than $0$ will follows.

YYF
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