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I need to determine the range of a that makes the iteration formula of matrix $\left(\begin{array}{lll} 1 & a & a \\ a & 1 & a \\ a & a & 1 \end{array}\right)$ converge when using Jacobian iteration method $\color{Gray} {\text{(2001 武汉 岩石 数值分析 7)}} $.

Reduce[And @@ Thread[Abs[Eigenvalues[-{{0, a, a}, {a, 0, a}, {a, a, 0}}]] < {1, 1, 1}], Reals]

In fact, this problem can also be solved by the following method:

Reduce[PositiveDefiniteMatrixQ[{{1, a, a}, {a, 1, a}, {a, a, 1}}] == 
   True && 
     PositiveDefiniteMatrixQ[
    2*IdentityMatrix[3] - {{1, a, a}, {a, 1, a}, {a, a, 1}}] == 
   True, a]

I want to know how to solve the above matrix equation?

The results of the following methods are correct, but the process is not exquisite and does not meet the requirements:

Reduce[Det[{{1, a}, {a, 1}}] > 0 && 
  Det[{{1, a, a}, {a, 1, a}, {a, a, 1}}] > 0 && 
     Det[2*IdentityMatrix[3] - {{1, a, a}, {a, 1, a}, {a, a, 1}}] > 
   0 && 
     Det[{{1, -a}, {-a, 1}}] > 0, a]
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    $\begingroup$ Is your question about how to change Abs[x] < 1/2 into -1/2 < x < 1/2? If so you might try with pattern matching: Abs[x] < 1/2 /. Abs[x_] < y_ :> -y < x < y. $\endgroup$ – anderstood Sep 15 at 7:38
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    $\begingroup$ Once again a question about the underlying math rather than the software Mathematica. Such questions should be directed to an appropriate forum, which this is not. $\endgroup$ – Daniel Lichtblau Sep 15 at 13:32
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First, the matrix m is positive definite of all the determinant and sub-determinants are > 0.

Second the Jacobi method will converge if the matrix is strictly diagonally dominant. That means in each row, the sum of the absolute values of the non diagonal elements is smaller than the absolute value of the diagonal element. In this case: 2 Abs[a] < 1

With this:

m = {{1, a, a}, {a, 1, a}, {a, a, 1}}
eq = {2 Abs[a] < 1, Det[m[[1 ;; 2, 1 ;; 2]]] > 0, Det[m] > 0};
Reduce[eq, a, Reals]
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