[proofplan] We apply the [Householder-John Theorem](/theorems/1390) with the Jacobi splitting $B = D$. The iteration matrix is $H_J = I - D^{-1}A$. The two hypotheses are: (1) $A$ is SPD (given), and (2) $B + B^\top - A = 2D - A$ is SPD (given by the second hypothesis of the theorem). These match the Householder-John conditions exactly, yielding $\rho(H_J) < 1$. [/proofplan] [step:Identify the Jacobi splitting and compute $B + B^\top - A$] The Jacobi method uses the splitting $B = D = \operatorname{diag}(a_{11}, \ldots, a_{nn})$. Since $B = D$ is symmetric, $B^\top = D$, and \begin{align*} B + B^\top - A = D + D - A = 2D - A. \end{align*} [guided] Recall that in any stationary iterative method with splitting $A = B - (B - A)$, the iteration matrix is $H = I - B^{-1}A$ and the iteration is $x^{(k+1)} = Hx^{(k)} + B^{-1}b$. For the Jacobi method specifically, $B$ is the diagonal part of $A$, so $B = D = \operatorname{diag}(a_{11}, \ldots, a_{nn})$. Since $D$ is a diagonal matrix, it is trivially symmetric: $D^\top = D$. The quantity $B + B^\top - A$ that appears in the Householder-John theorem therefore simplifies to $D + D - A = 2D - A$. This is a concrete, easily computable matrix whose positive definiteness can be checked for any given $A$. [/guided] [/step] [step:Verify the two hypotheses of the Householder-John Theorem] The [Householder-John Theorem](/theorems/1390) requires two conditions: 1. **$A$ is symmetric positive definite.** This is given by hypothesis. 2. **$B + B^\top - A = 2D - A$ is symmetric positive definite.** This is also given by hypothesis. Both conditions are satisfied. We also verify that $B = D$ is nonsingular (required for the splitting to define an iteration): since $A$ is SPD, its diagonal entries satisfy $a_{ii} = e_i^\top A e_i > 0$ for each $i$, so $D = \operatorname{diag}(a_{11}, \ldots, a_{nn})$ is invertible. [guided] We must check that the hypotheses of the Householder-John theorem are met before invoking it. The theorem requires (i) $A$ is SPD, and (ii) $B + B^\top - A$ is SPD. Condition (i) is a direct hypothesis of the theorem we are proving. Condition (ii) is the second hypothesis of the theorem we are proving, stated as "$2D - A$ is SPD." Note that $2D - A$ is automatically symmetric when $A$ is symmetric: $(2D - A)^\top = 2D^\top - A^\top = 2D - A$. So the SPD condition on $2D - A$ only requires verifying positive definiteness, i.e., $x^\top(2D - A)x > 0$ for all $x \neq 0$. We also need $D$ to be nonsingular so that the iteration matrix $H_J = I - D^{-1}A$ is well-defined. Since $A$ is SPD, each diagonal entry $a_{ii} = e_i^\top A e_i > 0$ (the quadratic form of an SPD matrix evaluated at any nonzero vector is positive, and the standard basis vector $e_i$ is nonzero). Therefore $D$ has all positive diagonal entries and is invertible. [/guided] [/step] [step:Conclude convergence of the Jacobi method] By the [Householder-John Theorem](/theorems/1390), $\rho(H_J) < 1$, where $H_J = I - D^{-1}A = I - B^{-1}A$ is the Jacobi iteration matrix. By the [Spectral Radius Criterion](/theorems/1387), $\rho(H_J) < 1$ implies $H_J^k z \to 0$ for every $z \in \mathbb{R}^n$. Since the error at step $k$ satisfies $e^{(k)} = H_J^k e^{(0)}$ by the [Error Propagation](/theorems/1386) theorem, $e^{(k)} \to 0$ as $k \to \infty$, and the Jacobi method converges for any initial vector $x^{(0)}$. [guided] The Jacobi case of the Householder-John theorem is particularly transparent: since $B = D$ is symmetric, $B + B^\top - A = 2D - A$, and the theorem reduces to the statement that $A$ SPD and $2D - A$ SPD together imply Jacobi convergence. What does the condition "$2D - A$ is SPD" mean? Writing $A = D + L_0 + U_0$ (with $U_0 = L_0^\top$ by symmetry), $2D - A = D - L_0 - U_0$. This is SPD when the diagonal $D$ "dominates" the off-diagonal part $L_0 + U_0$ in the spectral (quadratic form) sense: $x^\top Dx > x^\top(L_0 + U_0)x$ for all $x \neq 0$. This is weaker than strict diagonal dominance (which requires row-by-row domination $|a_{ii}| > \sum_{j \neq i} |a_{ij}|$), so the present result applies to some matrices where the [diagonal dominance criterion](/theorems/1389) does not. However, it requires $A$ to be SPD, which the diagonal dominance criterion does not. Unlike Gauss-Seidel for SPD matrices (which always converges by the [Gauss-Seidel Convergence for SPD Matrices](/theorems/1391) theorem), the Jacobi method requires the *additional* condition $2D - A$ SPD. There exist SPD matrices for which Jacobi diverges — for instance, certain tridiagonal SPD matrices with large off-diagonal entries can have $\rho(H_J) > 1$. The chain of reasoning is: Householder-John gives $\rho(H_J) < 1$; the spectral radius criterion converts this to $H_J^k \to 0$; and error propagation connects $H_J^k \to 0$ to convergence of the iterates $x^{(k)} \to A^{-1}b$. [/guided] [/step]