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Let the following $4 \times 4$ LGS be given for which all coefficients $a_1, a_2, a_3, a_{11}, a_{12}, ..., a_{33}$ are $>0$:

$a_1 + a_{11} \; x_1 + a_{12} \; x_2 + a_{13} \; x_3 + 0 \; x_4 = (a_{11}+a_{21}+a_{31}+a_{41}) \; x_1$ $a_2 + a_{21} \; x_1 + a_{22} \; x_2 + a_{23} \; x_3 + 0 \; x_4 =(a_{12}+a_{22}+a_{32}+a_{42}) \; x_2$ $a_3 + a_{31} \; x_1 + a_{32} \; x_2 + a_{33} \; x_3 + 0 \; x_4 =(a_{13}+a_{23}+a_{33}+a_{43})x_3 $
$a_4 + a_{41} \; x_1 + a_{42} \; x_2 + a_{43} \; x_3 + 0 \; x_4 =a_{44} \; x_4 $

and the corresponding matrix:

$A=\begin{pmatrix} -a_{21}-a_{31}-a_{41} & a_{12} & a_{13} & 0 \\ a_{21} & -a_{12}-a_{32}-a_{42} & a_{23} & 0 \\ a_{31} & a_{32} & -a_{13}-a_{23}-a_{43} & 0 \\ a_{41} & a_{42} & a_{43} & -a_{44} \\ \end{pmatrix}$

then:

1) The determinant is not equal to $0$ (i.e. there is exactly one solution).

2) The solution is positive (all elements of the vector are $> 0$ )

My question: Can this be generalized to any $n \times n$ LGS ( $\geq4$ ) ?

best regards carlo

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  • $\begingroup$ What about the signs of the coefficients $a_{41}\dots a_{44}$? What do you mean by "the solution is positive"? Is there a given right-hand side? $\endgroup$
    – Bertoldo Baccalà
    Commented Sep 25 at 13:00
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    $\begingroup$ What is an LGS, please? $\endgroup$
    – Gerry Myerson
    Commented Sep 26 at 0:18
  • $\begingroup$ 1) linear system of equations = LGS 2) You can write the LGS in matrix form: $\begin{pmatrix} -a_{21}-a_{31}-a_{41} & a_{12} & a_{13} & 0 \\ a_{21} & -a_{12}-a_{32}-a_{42} & a_{23} & 0 \\ a_{31} & a_{32} & -a_{13}-a_{23}-a_{43} & 0 \\ a_{41} & a_{42} & a_{43} & -a_{44} \\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{pmatrix}= \begin{pmatrix} -a_1\\ -a_2\\ -a_3\\ -a_4\\\end{pmatrix} $ \\ \\ \\ $\endgroup$
    – user508589
    Commented Sep 26 at 6:40
  • $\begingroup$ LGS stands for linear system of equations? In which language, please? $\endgroup$
    – Gerry Myerson
    Commented Sep 27 at 2:16
  • $\begingroup$ Question about the same system of equations on mathstack, math.stackexchange.com/questions/4977050/… $\endgroup$
    – Gerry Myerson
    Commented Sep 27 at 8:49

1 Answer 1

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Provided all the coefficients $a_{jk}$ you mentioned are positive, the answer is yes.

  1. Gershgorin's disk theorem applied to the upper left $(n-1)\times(n-1)$ submatrix shows that its eigenvalues are nonzero. Then Laplace's determinant formula teaches us that the determinant of $A$ is nonzero because $a_{nn}\neq0$.

  2. Assume the coefficients $a_1,\dots,a_n$ of the RHS are all positive. By the zeros in the last column, the system is reducible by block-Gaussian elimination. Since the upper left $(n-1)\times(n-1)$ matrix is monotone, the coefficients $x_1,\dots,x_{n-1}$ will be positive. Now, you can solve for $x_n$ explicitly and it will necessarily be positive.

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  • $\begingroup$ Thank you for your answer. I only require that all $a_{ij} >0$ and all $a_i > 0$ . Questions: 1) Are then all $x_i > 0$? 2) If not, is there a counterexample? best regards carlo $\endgroup$
    – user508589
    Commented Sep 25 at 18:28
  • $\begingroup$ The inverse of $-A$ is positive iff that matrix is monotone, see en.m.wikipedia.org/wiki/Monotone_matrix . By setting $a_{44}$ very small and, say, $a_{41}$ large, you can easily violate this property. $\endgroup$
    – Bertoldo Baccalà
    Commented Sep 25 at 19:13
  • $\begingroup$ $\begin{pmatrix} -a_{21}-a_{31}-a_{41} & a_{12} & a_{13} & 0 \\ a_{21} & -a_{12}-a_{32}-a_{42} & a_{23} & 0 \\ a_{31} & a_{32} & -a_{13}-a_{23}-a_{43} & 0 \\ a_{41} & a_{42} & a_{43} & -a_{44} \\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{pmatrix}= \begin{pmatrix} -a_1\\ -a_2\\ -a_3\\ -a_4\\\end{pmatrix} $ I set all $a_i$ and all coefficients $a_{ij}$ to 1, except $a_{41}$ and $a_{44}$: \ $a_{41} = 1000$ (large) and $t_4 = -1/1000$ (very small). But Wolfram Alpha only provides a positive solution (all $x_i > 0$) $\endgroup$
    – user508589
    Commented Sep 26 at 6:57
  • $\begingroup$ Due to the special structure, you can indeed show that the solution is always positive. I have revised the above answer accordingly. $\endgroup$
    – Bertoldo Baccalà
    Commented Sep 26 at 8:35
  • $\begingroup$ What ist the difference between "block-Gaussian elimination" and "Gaussian algorithmus" ? Why ist the upper left $n-1$ x $n-1$ matrix monotone? I can only prove that it has the Determinant unequal to 0. best regards carlo $\endgroup$
    – user508589
    Commented Sep 26 at 13:16

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